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What is the characteristic equation form for `y'' - 6y' + 9y = 0`?
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The characteristic equation is `r^2 - 6r + 9 = 0`.
Solve `y'' - 5y' + 6y = 0` by finding the general solution.
The general solution is `y = C1 * e^(2x) + C2 * e^(3x)`.
When solving second-order linear differential equations, what is the related characteristic equation from `ay'' + by' + cy = 0`?
The characteristic equation is `ar^2 + br + c = 0`.
What happens when `b^2 - 4ac = 0` in the characteristic equation?
There is one real repeated root `r = -b / 2a`, and the general solution is `y = (C1 + C2x) * e^(rx)`.
Given the characteristic equation `r^2 - 5r + 6 = 0`, identify the roots.
`r = 2` and `r = 3`.
For `9y'' + 24y' + 16y = 0`, identify and solve the characteristic equation.
The characteristic equation is `9r^2 + 24r + 16 = 0`. Factoring gives `(3r + 4)^2 = 0`, so `r = -4/3`. The general solution is `y = (C1 + C2x) * e^(-4x/3)`.
What is the general solution for the differential equation when the characteristic equation has two distinct real roots?
The general solution is `y = C1 * e^(r1x) + C2 * e^(r2x)`, where `r1` and `r2` are the distinct roots.
Solve for the roots and general solution of `y'' + 8y' + 25y = 0`.
The characteristic equation is `r^2 + 8r + 25 = 0`. Solving using the quadratic formula gives `r = -4 ± 3i`. The general solution is `y = e^(-4x) * (C1*cos(3x) + C2*sin(3x))`.
How do you solve the boundary value problem `y'' - 2y' + y = 0` with `y(0) = 3` and `y(1) = 7e`?
First, solve the characteristic equation: `(r - 1)^2 = 0`. General solution: `y = (C1 + C2x) * e^x`. Using boundary conditions: `y(0) = 3` gives `C1 = 3`, and `y(1) = 7e` gives `3e + 4e = 7e`, so `C2 = 4`. The particular solution is `y = (3 + 4x) * e^x`.
Describe the roots and general solution when `b^2 - 4ac < 0` for the characteristic equation.
The roots are complex in the form `r = α ± βi`, and the general solution is `y = e^(αx) * (C1 * cos(βx) + C2 * sin(βx))`.
Given `y'' + 4y = 0` with initial conditions `y(0) = 4` and `y'(0) = 6`, find the particular solution.
The general solution is `y = C1 * cos(2x) + C2 * sin(2x)`. Using initial conditions: `y(0) = 4` gives `C1 = 4`, and `y'(0) = 6` gives `2C2 = 6`, so `C2 = 3`. The particular solution is `y = 4 * cos(2x) + 3 * sin(2x)`.
What are the general forms of second-order linear differential equations?
They are of the form `y'' + p(x)y' + q(x)y = g(x)`, where `p(x)`, `q(x)`, and `g(x)` are continuous functions. If `g(x) = 0`, it is homogeneous; otherwise, it is non-homogeneous.
What are the roots of the characteristic equation when `b^2 - 4ac > 0`?
When `b^2 - 4ac > 0`, there are two distinct real roots, which can be found using the quadratic formula `r = (-b ± sqrt(b² - 4ac)) / 2a`.
What is the general solution of `y'' - 6y' + 9y = 0`?
The general solution is `y = (C1 + C2x) * e^(3x)` since the root `r = 3` is repeated.
How do you recognize a differential equation with constant coefficients?
It will be in the form `ay'' + by' + cy = 0` where `a, b, and c` are constants.
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