Eliminating Constants in Differential Equations

Hi, so for today we're going to talk about the elimination of arbitrary constants and this is still a part of a differential equation and it is an important topic in differential equations because upon eliminating the arbitrary constants here, arbitrary constants here in our equations, we will be able to determine the differential equation of that certain equation and it is also useful. This topic is also useful actually if you are already on the study of the families of curves which is actually the next lesson after elimination of arbitrary constants. So the primary goal here is you are given an equation and you're going to eliminate the constants in order for you to obtain the DE. So let's get started. For today we have seven examples and we shall be getting the differential equation by eliminating the arbitrary constants for each of the equations so let's start for number one we have y is equals to ax squared plus bx plus c so our goal here is to eliminate the arbitrary constants to obtain the DE so for number one we have y is equals to ax squared plus bx plus c. So number one, you need to determine how many constants are present in the equation. So it basically tells us how many times we are going to derive the function. So in this case, we have the constants we know. We have the constants a, b, and c. So we're going to get the third derivative. It means that we're going to derive this function c. three times. The number of constant defines how many times we are going to derive the function. So for example, so we're going to get the y prime of this. So what will happen here is 2ax plus b and plus 0 because the derivative of this with respect to x, this is a constant, this is 0. So in other words, we have a y prime that is equal to 2ax plus b. The second derivative of that would be 2a plus 0 because the derivative of this constant b is again 0. So in other words, we have y double prime 2a only. And of course, if we're going to get the third derivative of that, we shall be getting a 0. So the differential equation for this equation or function y is equal to ax squared plus bx plus c is equal only to the y triple prime is equal to zero or the third derivative with respect to x is equal to zero that is the differential equation we have obtained so for our next example we have number two we have x cubed minus 3x squared y is equal to c so oftentimes the differential equation uses the variable c to denote that this that is a that is an arbitrary constant so in this case we have to determine how many constants is present in the equation. So in this case we have we only have one constant and that is denoted by this letter C, a constant. So it means that if the arbitrary constant is only one we need to differentiate the function or the whole equation only once. So differentiating this we have 3 x squared minus 3. This is a product rule so you have to get copy first the first variable and then get the derivative of this so we have y prime plus copy this and get the derivative of this that is 2x okay equals to zero because the derivative of constant is zero thus we have eliminated already the arbitrary constant and all we have to do is to simplify this equation in order for us to get the differential equation so we have 3x squared minus 3x squared y prime minus 6x y is equal to 0 and we can further simplify this by actually dividing this by 1 over 3x again as you can see 3 0 will cancel we have 3 and 1 term of X will cancel so this X will be cancelled this will become 2 so what will happen we have X minus Y or I mean X to Y prime okay so we have x y prime minus 2y is equals to 0 and take note that y prime is equal to dy over d in this case so we can further rewrite this as x times the dy over dx minus 2y is equal to 0 and furthermore we can actually multiply the whole equation by dx so that we have x dx minus x dy minus 2y dx is equal to 0 and combining the terms which have dx, okay, we have this equation now is equal to zero, but we can factor out x minus 2y, okay, or the dx here, so that our differential equation would look like this. So this would be our differential equation produced upon eliminating the arbitrary constant, which is this constant. Okay, so the number of arbitrary constant again defines how many times do we have to differentiate the given equation. So let's continue. For number 3, we have the problem. Number 3, we have y is equal to c sub 1, constant 1, plus c sub 2. Oops, this is c sub 2. c sub 2 times e raised to 3x. It is very evident that in this equation, we have two arbitrary constants. So it means that we have to differentiate this y twice. So let's get started. y prime, the derivative of this constant is 0. You get the derivative of this. So we have 3 c sub 2 e raised to 3x. And Get still, the second derivative of this, because we have two constants, as I have said a while ago, so this should be 9 C sub 2 E raised to 3x. So in other words, let me just simplify this. So we have y'equals to 3 C sub 2 E raised to 3 raised to x, C sub 2 E raised to 3x, and we have 9 C sub 2 E raised to 3x. So what will happen? So we have to eliminate C sub 2. So this can be our equation 1 and this can be our equation 2 and we have to eliminate the c sub 2. So from 1 and 2, what are we going to do is to write the equation. We have y prime is equal to 3 c sub 2 e raised to 3x and we have y double prime is equal to 9 c sub 2 e raised to 3x. So I can multiply this first equation by 3. by 3 I can multiply that and what will we do is to subtract it from the other from the second equation so if I'm going to multiply this by 3 we have 3 y prime is equals to 9 c sub 2 e raised to 3 x okay and we have y double prime 9 c sub 2 e raised to 3 x and if I'm going okay If I'm going to negate this, okay, multiply this by negative 1, okay, and add, what will happen is this. So we have negative 3y'equals to negative 9c sub 2e3 raised to e3x plus we have the y double prime equals to 9c sub 2e3x, okay. So what will happen here is that These two will cancel if I'm going to add these two. Evidently, this will be equal to zero. And we have now the differential equation by we have negative 3y prime plus y double prime is equal to zero. Or in other words, we have y double prime minus 3y prime is equal to zero. This should be our differential equation. So let's try to solve another problem. Problem number four. So we have here for number 4, we have here y, y sine of x minus x y squared is equal to c. It is evident also here in this equation that we have already one arbitrary constant, which is c. So we're going to find the derivative once. this full equation so that we are going to eliminate the arbitrary constant C. So first, this is actually a product rule so what will happen here is that we have Y okay Y then the derivative of this is actually cosine of X plus okay Y cosine of X then what's next we have to get the derivative of this Y so that in this case that is dy over the X okay or Y prime or Y prime then copy the sine of x minus okay still this is the uh a product rule so what will happen here we have x okay derivative of this we have 2y y prime or 2y dy over dx plus we have the y squared multiplied by 1 equal to 0 1 because the derivative of x is 1 so we have here y cosine of x plus sine of x, dy multiplied by dy over dx minus 2xy. Let's say we have dy over dx also of y prime minus we have dy squared equals to 0. And what are we going to do for this equation? We have to eliminate the dx. So we multiply. All equation by the dx, y cosine of x dx plus we have sine of x dy minus 2xy dy minus y squared dx and that is equivalent to 0, equal to 0. So as you can see here, the arbitrary constant is no longer present because we differentiate it already. So what are we going to do here is the same process as what we have done a while ago. We have to combine them in order for us to factor the dx. So y cosine of x dx minus y squared dx plus sine of x dy minus 2xy dy is equals to 0. So factoring out dx, we have y cosine of x minus y squared. dx okay dx and then we have plus sine of x minus 2xy dy is equals to 0 or in other words still here we can still factor out y so we have y we have y cosine of x minus y dx plus a sine of x minus 2x to y dy is equals to 0. So this should be our differential equation. Okay, our differential equation for the given example. Thus, we have eliminated already the arbitrary constant, which is the main goal of this video, to eliminate those arbitrary constants. Okay, let's go now to problem number 5. So we have problem number 5. Number 5 is given as x equals to, we have c sub 1 cosine omega t plus c sub 2 the sine of omega t. Okay, as you can see here, our equation contains how many arbitrary constants? Well, that's correct. That's 2. Okay, we have C sub 1 and C sub 2. So it means that we're going to get the derivative of this x with respect to t to eliminate the C sub 1 and the C sub 2 constant. So what are we going to do here? Is that in order for us to eliminate that, we have to get dx over dt. So what will happen? So well, of course, if we're going to derivative this, this is a constant. so we cannot easily eliminate this constant because as we all know if this is a constant let's say two or one if we're going to get the derivative of this we have to apply chain rule and definitely this C sub 1 still will not be eliminated okay so but still the same rule applies we have two arbitrary constants so what are we going to do is still differentiate okay so upon differentiating this the derivative of the inside function with respect to t is actually omega. C sub 1 the derivative of cosine is negative sign of which I'm going to write the negative sign here okay sine omega t agree or not yes that's correct plus still omega get the derivative of this still omega then the derivative of sine omega t is uh let's say still we have d constant here this would be cosine omega t okay that is still omega cosine omega t So this is the first derivative as you can see it is not evident still that the c1 and c2 are canceled but definitely we will find a way on how are we going to eliminate that by simply getting the derivative so hence we now get the we can still now get the second derivative of x with respect to t of this function so what will happen is that same process so derivative of this is w multiplied by the negative double here, this would be negative double squared, C sub 1. Still, the presence of C sub 1 is still there. Sine omega t, derivative of that is cosine omega t. Plus, so what will happen here? Still the same, derivative of the inside function of the cosine is omega times omega, that is omega squared, C sub 2. And the derivative of the whole trigonometric function is negative sine omega t. So what will happen is that negative omega squared c sub 1 cosine omega t minus omega squared c sub 2. Oops, let me just rewrite it so that it would look like a c sub 2. We have c sub 2 sine omega t. Okay, here's the thing. So as we all know, we can factor out the negative omega squared outside of this. secondary derived function so we have negative omega squared so we are left with this is negative so it should be positive c sub 1 cosine omega t you get negative you show this should be plus c sub 2 sine omega t and actually this whole function this whole term is actually equal to our x which is what which is this So we can simply substitute or eliminate the c sub 1 and c sub 2 by substituting this. Okay? This term into x, simply x. So what will happen? So we have the second derivative of x with respect to t. And now we have negative omega squared x. So that is our differential equation. Rearranging to eliminate the negative sign. So what will happen is this. So we have second derivative of x with respect to t plus omega squared x is equal to 0. So this should be our differential equation. Okay, let's proceed with the number 6. Okay, so we have number 6. We have 7 examples here. So last two, go. So we have x is equal to a sine omega t plus b, okay? Where a and b here are constants. So we can have a derivative, the second derivative of this, in order for us to eliminate the arbitrary constant. So the same process, dx with respect to t, of course. So what will happen is with respect to t, we're going to derive a... Take the derivative of this inside function. Okay, with respect to t, it would be omega. This is a constant, so that would be 0. So we have omega, okay, a. Derivative of sine omega t plus b is cosine omega t plus b. Or we can say a omega cosine omega t plus b. Okay, so getting the second derivative, of course, with respect to t. Still, so what will happen? What will happen is chain rule again. So we have negative a omega squared negative because the derivative of cosine is negative sine. So we have sine omega t plus b. And as you can see, this is a very similar problem to what we have a while ago in the previous problem. So we derive, we differentiate this function and still. there is the presence of arbitrary constants so what we're going to do this is second derivative x with respect to t so we can factor out omega squared and we have a sine omega t plus b which is actually equal to our value of still x okay so from the equation here above so simply we have the second derivative of x with respect to t that is equivalent to negative Omega squared x and rearranging the equation, we now have second derivative of x with respect to e plus omega squared x equals to 0. So that is our differential equation. As you can see, that is still the same answer, same differential equation as we have from the previous example. And down for our last example, which is number 7. So number 7, we have y is equal to c sub 1 e raised to negative 2x plus c sub 2 e raised to 3x. It's evident that in this equation, we have two arbitrary constants. Thus, we have to perform differentiation twice. So we have y prime. Negative 2, c sub 1, e raised to negative 2x, plus 3, c sub 2, e raised to 3x. Okay? So this should be our, this is our first equation. Let's say that this is our first equation. So this is our second equation. For y double prime, we have, okay, the derivative of that is 4, c sub 1, e raised to negative 2x. 4 because the derivative of this, again, is negative 2, negative 2. times negative 2 should yield to 4 and still copy the same function. So we have plus 9c sub 2 e raised to 3x. So we can label this as equation. So as you can see here, we will be using systems of linear equation in order for us to eliminate the arbitrary constants. So let's try. So from equation 2 and 3, okay? From equation 2, and 3 let's try to eliminate some let's try to eliminate the c sub 1 okay let's try to eliminate c sub 1 so from equation 2 we have y prime equals to negative 2 c sub 1 e raised to negative 2x plus 3 c sub 2 e raised to 3x and why did i say that we can eliminate c sub 1 easily it's because if we're going to multiply this okay by 2 and add we can simply cancel this term so we have 9c sub 2 e raised to 3x so we're going to multiply this by 2 okay and what will happen here is that we have 2y prime is equals to negative 4 c sub 1 e raised to negative 2x plus 2 times 3 that is that is 6 c sub 2 e raised to 3x okay And we're going to plus or add this in our third equation, which is y double prime is equal to 4c sub 1 e raised to negative 2x. And simply this would be cancelled. Cancelled, okay? That should be equal to 0. And this is plus 9c sub 2 e raised to 3x. So what will happen is that upon adding these two equations, so we have now y double prime plus 2y prime, okay? equals to 15 C sub 2 e raised to 3x. Okay let's label this as our equation 4. And again still we're going to use equation the other equation 1 and 2 we're going to find a way on how are we going to cancel C sub 1 again so that we are only left with C sub 2. So in this case from the equation 1 and 2 we're going to get equation 1 Equation 1 and equation 2. So we have, what's our equation 1? The original function, y equals to c sub 1 e raised to negative 2x plus c sub 2 e raised to 3x. And our equation 2 is y prime. Okay, again, y prime. Negative 2, c sub 1, basically the derivative of this, negative 2, c sub 1 e raised to negative 2x plus 3C sub 2E raised to 3X. So we can easily eliminate the C sub 1 here by simply multiplying this first equation. Okay. We're going to multiply it by, again, by 2. So what will happen here? We have 2Y is equal to 2C sub 1E raised to negative 2X plus 2C sub 2E raised to 3X. And we're going to add... We're going to add that to our second equation which is y prime negative 2 c sub 1 e raised to negative 2 x plus 3 c sub 2 e raised to 3 x. So this would cancel and we have y prime plus 2 y is equals to 5 c sub 2 e raised to 3 x and that yields to our another equation equation 5. So as you can see we now eliminate the c sub 1 and it's time to eliminate c sub 2 so from equation 4 and 5 okay we're going to try to eliminate c sub 2 so we have y double prime for equation 4 plus 2y prime is equals to 15 c sub 2e raised to 3x and from equation 5 we have y prime plus 2y is equal to 5c sub 2e raised to 3x. So the wonderful thing here is that we're going to multiply this whole equation, let's say, by negative 3, and we're going to add it to our equation 4. What will happen? So we have negative 3y prime minus 6y. is equal to negative 15 c sub 2 e raised to 3x. So we're going to add to the original equation of equation 4 we have y double prime plus 2 y prime is equal to 15 c sub 2 e raised to 3x. So as you can see if we're going to add these two this would cancel eliminating the c sub 2 and that is our primary goal to eliminate the arbitrary constant. So that what will happen here we're going to add So y double prime is still y double prime. 2y prime minus 3y prime plus negative 3y prime is negative y prime. And plus this should be 6y is equal to 0. So as you can see here, this is our differential equation. And that is all for today. That is the elimination of arbitrary constants. And the wonderful thing to remember here is that we have to remember that we have to have a differential equation. to note how many arbitrary constants are present in the equation and you're going to differentiate it based on how many arbitrary constants is present in the equation and you're going to find a way to eliminate the arbitrary constant. So the arbitrary constant cannot be sometimes be eliminated by simply just differentiating. You have to find ways in order for you to eliminate the arbitrary constant. If you find this video helpful, thank you so much. Please subscribe to my channel and this is Engineer Abad and thank you so much for listening and God bless you.

So let's get started. For today we have seven examples and we shall be getting the differential equation by eliminating the arbitrary constants for each of the equations so let's start for number one we have y is equals to ax squared plus bx plus c so our goal here is to eliminate the arbitrary constants to obtain the DE so for number one we have y is equals to ax squared plus bx plus c. So number one, you need to determine how many constants are present in the equation. So it basically tells us how many times we are going to derive the function. So in this case, we have the constants we know.

We have the constants a, b, and c. So we're going to get the third derivative. It means that we're going to derive this function c. three times.

The number of constant defines how many times we are going to derive the function. So for example, so we're going to get the y prime of this. So what will happen here is 2ax plus b and plus 0 because the derivative of this with respect to x, this is a constant, this is 0. So in other words, we have a y prime that is equal to 2ax plus b. The second derivative of that would be 2a plus 0 because the derivative of this constant b is again 0. So in other words, we have y double prime 2a only.

And of course, if we're going to get the third derivative of that, we shall be getting a 0. So the differential equation for this equation or function y is equal to ax squared plus bx plus c is equal only to the y triple prime is equal to zero or the third derivative with respect to x is equal to zero that is the differential equation we have obtained so for our next example we have number two we have x cubed minus 3x squared y is equal to c so oftentimes the differential equation uses the variable c to denote that this that is a that is an arbitrary constant so in this case we have to determine how many constants is present in the equation. So in this case we have we only have one constant and that is denoted by this letter C, a constant. So it means that if the arbitrary constant is only one we need to differentiate the function or the whole equation only once. So differentiating this we have 3 x squared minus 3. This is a product rule so you have to get copy first the first variable and then get the derivative of this so we have y prime plus copy this and get the derivative of this that is 2x okay equals to zero because the derivative of constant is zero thus we have eliminated already the arbitrary constant and all we have to do is to simplify this equation in order for us to get the differential equation so we have 3x squared minus 3x squared y prime minus 6x y is equal to 0 and we can further simplify this by actually dividing this by 1 over 3x again as you can see 3 0 will cancel we have 3 and 1 term of X will cancel so this X will be cancelled this will become 2 so what will happen we have X minus Y or I mean X to Y prime okay so we have x y prime minus 2y is equals to 0 and take note that y prime is equal to dy over d in this case so we can further rewrite this as x times the dy over dx minus 2y is equal to 0 and furthermore we can actually multiply the whole equation by dx so that we have x dx minus x dy minus 2y dx is equal to 0 and combining the terms which have dx, okay, we have this equation now is equal to zero, but we can factor out x minus 2y, okay, or the dx here, so that our differential equation would look like this. So this would be our differential equation produced upon eliminating the arbitrary constant, which is this constant.

Okay, so the number of arbitrary constant again defines how many times do we have to differentiate the given equation. So let's continue. For number 3, we have the problem.

Number 3, we have y is equal to c sub 1, constant 1, plus c sub 2. Oops, this is c sub 2. c sub 2 times e raised to 3x. It is very evident that in this equation, we have two arbitrary constants. So it means that we have to differentiate this y twice.

So let's get started. y prime, the derivative of this constant is 0. You get the derivative of this. So we have 3 c sub 2 e raised to 3x.

And Get still, the second derivative of this, because we have two constants, as I have said a while ago, so this should be 9 C sub 2 E raised to 3x. So in other words, let me just simplify this. So we have y'equals to 3 C sub 2 E raised to 3 raised to x, C sub 2 E raised to 3x, and we have 9 C sub 2 E raised to 3x. So what will happen?

So we have to eliminate C sub 2. So this can be our equation 1 and this can be our equation 2 and we have to eliminate the c sub 2. So from 1 and 2, what are we going to do is to write the equation. We have y prime is equal to 3 c sub 2 e raised to 3x and we have y double prime is equal to 9 c sub 2 e raised to 3x. So I can multiply this first equation by 3. by 3 I can multiply that and what will we do is to subtract it from the other from the second equation so if I'm going to multiply this by 3 we have 3 y prime is equals to 9 c sub 2 e raised to 3 x okay and we have y double prime 9 c sub 2 e raised to 3 x and if I'm going okay If I'm going to negate this, okay, multiply this by negative 1, okay, and add, what will happen is this.

So we have negative 3y'equals to negative 9c sub 2e3 raised to e3x plus we have the y double prime equals to 9c sub 2e3x, okay. So what will happen here is that These two will cancel if I'm going to add these two. Evidently, this will be equal to zero. And we have now the differential equation by we have negative 3y prime plus y double prime is equal to zero. Or in other words, we have y double prime minus 3y prime is equal to zero.

This should be our differential equation. So let's try to solve another problem. Problem number four. So we have here for number 4, we have here y, y sine of x minus x y squared is equal to c. It is evident also here in this equation that we have already one arbitrary constant, which is c.

So we're going to find the derivative once. this full equation so that we are going to eliminate the arbitrary constant C. So first, this is actually a product rule so what will happen here is that we have Y okay Y then the derivative of this is actually cosine of X plus okay Y cosine of X then what's next we have to get the derivative of this Y so that in this case that is dy over the X okay or Y prime or Y prime then copy the sine of x minus okay still this is the uh a product rule so what will happen here we have x okay derivative of this we have 2y y prime or 2y dy over dx plus we have the y squared multiplied by 1 equal to 0 1 because the derivative of x is 1 so we have here y cosine of x plus sine of x, dy multiplied by dy over dx minus 2xy.

Let's say we have dy over dx also of y prime minus we have dy squared equals to 0. And what are we going to do for this equation? We have to eliminate the dx. So we multiply. All equation by the dx, y cosine of x dx plus we have sine of x dy minus 2xy dy minus y squared dx and that is equivalent to 0, equal to 0. So as you can see here, the arbitrary constant is no longer present because we differentiate it already. So what are we going to do here is the same process as what we have done a while ago.

We have to combine them in order for us to factor the dx. So y cosine of x dx minus y squared dx plus sine of x dy minus 2xy dy is equals to 0. So factoring out dx, we have y cosine of x minus y squared. dx okay dx and then we have plus sine of x minus 2xy dy is equals to 0 or in other words still here we can still factor out y so we have y we have y cosine of x minus y dx plus a sine of x minus 2x to y dy is equals to 0. So this should be our differential equation.

Okay, our differential equation for the given example. Thus, we have eliminated already the arbitrary constant, which is the main goal of this video, to eliminate those arbitrary constants. Okay, let's go now to problem number 5. So we have problem number 5. Number 5 is given as x equals to, we have c sub 1 cosine omega t plus c sub 2 the sine of omega t. Okay, as you can see here, our equation contains how many arbitrary constants?

Well, that's correct. That's 2. Okay, we have C sub 1 and C sub 2. So it means that we're going to get the derivative of this x with respect to t to eliminate the C sub 1 and the C sub 2 constant. So what are we going to do here?

Is that in order for us to eliminate that, we have to get dx over dt. So what will happen? So well, of course, if we're going to derivative this, this is a constant.

so we cannot easily eliminate this constant because as we all know if this is a constant let's say two or one if we're going to get the derivative of this we have to apply chain rule and definitely this C sub 1 still will not be eliminated okay so but still the same rule applies we have two arbitrary constants so what are we going to do is still differentiate okay so upon differentiating this the derivative of the inside function with respect to t is actually omega. C sub 1 the derivative of cosine is negative sign of which I'm going to write the negative sign here okay sine omega t agree or not yes that's correct plus still omega get the derivative of this still omega then the derivative of sine omega t is uh let's say still we have d constant here this would be cosine omega t okay that is still omega cosine omega t So this is the first derivative as you can see it is not evident still that the c1 and c2 are canceled but definitely we will find a way on how are we going to eliminate that by simply getting the derivative so hence we now get the we can still now get the second derivative of x with respect to t of this function so what will happen is that same process so derivative of this is w multiplied by the negative double here, this would be negative double squared, C sub 1. Still, the presence of C sub 1 is still there. Sine omega t, derivative of that is cosine omega t.

Plus, so what will happen here? Still the same, derivative of the inside function of the cosine is omega times omega, that is omega squared, C sub 2. And the derivative of the whole trigonometric function is negative sine omega t. So what will happen is that negative omega squared c sub 1 cosine omega t minus omega squared c sub 2. Oops, let me just rewrite it so that it would look like a c sub 2. We have c sub 2 sine omega t.

Okay, here's the thing. So as we all know, we can factor out the negative omega squared outside of this. secondary derived function so we have negative omega squared so we are left with this is negative so it should be positive c sub 1 cosine omega t you get negative you show this should be plus c sub 2 sine omega t and actually this whole function this whole term is actually equal to our x which is what which is this So we can simply substitute or eliminate the c sub 1 and c sub 2 by substituting this.

Okay? This term into x, simply x. So what will happen? So we have the second derivative of x with respect to t. And now we have negative omega squared x.

So that is our differential equation. Rearranging to eliminate the negative sign. So what will happen is this. So we have second derivative of x with respect to t plus omega squared x is equal to 0. So this should be our differential equation. Okay, let's proceed with the number 6. Okay, so we have number 6. We have 7 examples here.

So last two, go. So we have x is equal to a sine omega t plus b, okay? Where a and b here are constants.

So we can have a derivative, the second derivative of this, in order for us to eliminate the arbitrary constant. So the same process, dx with respect to t, of course. So what will happen is with respect to t, we're going to derive a... Take the derivative of this inside function.

Okay, with respect to t, it would be omega. This is a constant, so that would be 0. So we have omega, okay, a. Derivative of sine omega t plus b is cosine omega t plus b.

Or we can say a omega cosine omega t plus b. Okay, so getting the second derivative, of course, with respect to t. Still, so what will happen?

What will happen is chain rule again. So we have negative a omega squared negative because the derivative of cosine is negative sine. So we have sine omega t plus b.

And as you can see, this is a very similar problem to what we have a while ago in the previous problem. So we derive, we differentiate this function and still. there is the presence of arbitrary constants so what we're going to do this is second derivative x with respect to t so we can factor out omega squared and we have a sine omega t plus b which is actually equal to our value of still x okay so from the equation here above so simply we have the second derivative of x with respect to t that is equivalent to negative Omega squared x and rearranging the equation, we now have second derivative of x with respect to e plus omega squared x equals to 0. So that is our differential equation. As you can see, that is still the same answer, same differential equation as we have from the previous example.

And down for our last example, which is number 7. So number 7, we have y is equal to c sub 1 e raised to negative 2x plus c sub 2 e raised to 3x. It's evident that in this equation, we have two arbitrary constants. Thus, we have to perform differentiation twice.

So we have y prime. Negative 2, c sub 1, e raised to negative 2x, plus 3, c sub 2, e raised to 3x. Okay? So this should be our, this is our first equation. Let's say that this is our first equation.

So this is our second equation. For y double prime, we have, okay, the derivative of that is 4, c sub 1, e raised to negative 2x. 4 because the derivative of this, again, is negative 2, negative 2. times negative 2 should yield to 4 and still copy the same function.

So we have plus 9c sub 2 e raised to 3x. So we can label this as equation. So as you can see here, we will be using systems of linear equation in order for us to eliminate the arbitrary constants. So let's try.

So from equation 2 and 3, okay? From equation 2, and 3 let's try to eliminate some let's try to eliminate the c sub 1 okay let's try to eliminate c sub 1 so from equation 2 we have y prime equals to negative 2 c sub 1 e raised to negative 2x plus 3 c sub 2 e raised to 3x and why did i say that we can eliminate c sub 1 easily it's because if we're going to multiply this okay by 2 and add we can simply cancel this term so we have 9c sub 2 e raised to 3x so we're going to multiply this by 2 okay and what will happen here is that we have 2y prime is equals to negative 4 c sub 1 e raised to negative 2x plus 2 times 3 that is that is 6 c sub 2 e raised to 3x okay And we're going to plus or add this in our third equation, which is y double prime is equal to 4c sub 1 e raised to negative 2x. And simply this would be cancelled. Cancelled, okay? That should be equal to 0. And this is plus 9c sub 2 e raised to 3x.

So what will happen is that upon adding these two equations, so we have now y double prime plus 2y prime, okay? equals to 15 C sub 2 e raised to 3x. Okay let's label this as our equation 4. And again still we're going to use equation the other equation 1 and 2 we're going to find a way on how are we going to cancel C sub 1 again so that we are only left with C sub 2. So in this case from the equation 1 and 2 we're going to get equation 1 Equation 1 and equation 2. So we have, what's our equation 1?

The original function, y equals to c sub 1 e raised to negative 2x plus c sub 2 e raised to 3x. And our equation 2 is y prime. Okay, again, y prime.

Negative 2, c sub 1, basically the derivative of this, negative 2, c sub 1 e raised to negative 2x plus 3C sub 2E raised to 3X. So we can easily eliminate the C sub 1 here by simply multiplying this first equation. Okay.

We're going to multiply it by, again, by 2. So what will happen here? We have 2Y is equal to 2C sub 1E raised to negative 2X plus 2C sub 2E raised to 3X. And we're going to add... We're going to add that to our second equation which is y prime negative 2 c sub 1 e raised to negative 2 x plus 3 c sub 2 e raised to 3 x.

So this would cancel and we have y prime plus 2 y is equals to 5 c sub 2 e raised to 3 x and that yields to our another equation equation 5. So as you can see we now eliminate the c sub 1 and it's time to eliminate c sub 2 so from equation 4 and 5 okay we're going to try to eliminate c sub 2 so we have y double prime for equation 4 plus 2y prime is equals to 15 c sub 2e raised to 3x and from equation 5 we have y prime plus 2y is equal to 5c sub 2e raised to 3x. So the wonderful thing here is that we're going to multiply this whole equation, let's say, by negative 3, and we're going to add it to our equation 4. What will happen? So we have negative 3y prime minus 6y.

is equal to negative 15 c sub 2 e raised to 3x. So we're going to add to the original equation of equation 4 we have y double prime plus 2 y prime is equal to 15 c sub 2 e raised to 3x. So as you can see if we're going to add these two this would cancel eliminating the c sub 2 and that is our primary goal to eliminate the arbitrary constant. So that what will happen here we're going to add So y double prime is still y double prime.

2y prime minus 3y prime plus negative 3y prime is negative y prime. And plus this should be 6y is equal to 0. So as you can see here, this is our differential equation. And that is all for today. That is the elimination of arbitrary constants.

And the wonderful thing to remember here is that we have to remember that we have to have a differential equation. to note how many arbitrary constants are present in the equation and you're going to differentiate it based on how many arbitrary constants is present in the equation and you're going to find a way to eliminate the arbitrary constant. So the arbitrary constant cannot be sometimes be eliminated by simply just differentiating. You have to find ways in order for you to eliminate the arbitrary constant. If you find this video helpful, thank you so much.

Please subscribe to my channel and this is Engineer Abad and thank you so much for listening and God bless you.

Transcript for:Eliminating Constants in Differential Equations

Transcript for:
Eliminating Constants in Differential Equations