solve d ² y by DX 2 - 7 Dy by DX - 44 44 y = 0 we find the complete solution means using a complimentary function okay solution now given equation is uh given equation is d ² y - 7 d y - 44 y = 0 d stand d by DX therefore d ² - 7 d - 44 into y = 0 okay now this factorization or okay first auxilary equation is its auxiliary equation is m² - 7 m - 44 = 0 now you this factorization now you get m + 4 into M - 11 = 0 r m is equal to 1 is a M1 is -4 and second one is M2 is a plus 11 two roots are different okay so M1 is - 4 M2 is 11 these two roots are different now solution is therefore its solution is complementary function is equal to C1 e to - 4x 4X means is a d Dy by DX 4x + C2 e the power 11x okay this is a required solution next example solve D 4- 5 D ² + 4 into y = 0 now D 4 means uh four or four roots required we need a four roots now solution so right direct uh axillary equation its axillary equation is M 4 - 5 m² + 4 = Z this is only auxiliary equation okay now factorization this one m to 4 - 4 m² - M m² + 4 = 0 now m² is common m² m² - 4 and this is a take -1 m² - 4 = 0 okay now these two same uh therefore m² - 4 into m² - 1 = 0 okay now again the first bracket is M - 2 into m + 2 a s - b s a - b into a + b okay so m s - 4 is a m - 2 into m + 2 same same is a square - b square formula M -1 into m + 1 = to0 okay now roots are the first M1 is equal to First root m = 2 and M + 2 = 0 or M = -2 and M - 1 = 0 m = + 1 and this is a -1 the roots are different now complementary function is the roots are different C1 e 2x + C2 e - 2x + C3 into e to x + C4 e - x so required complimentary function okay next is next example solve D Cub y by DX [Music] CU - 2 d ² y by DX 2 - 4 4 Dy by DX + 8 y = 0 we need a complete solution means complimentary function now solution First I write the given equation is D Cub - 2 d ² 2 - 4 d + 8 into Y is a common equal to z d stand d by DX in this example now write auxiliary equation is it's auxiliary equation is M Cub - 2 m² - 4 m + 8 = 0 okay now see the first two terms third and four terms in this m sare is common no need for factorization so M - 2 first two terms the remaining - 4 is common again M - 2 equal to Z okay now first is M - 2 bracket is common then remaining into m² - 4 = 0 now therefore M - 2 into a s - B squ formula a - b into a + b means a - b into a + b m - 2 into m + 2 = 0 okay see next p continue now therefore c m - 2 m -2 m + 2 now roots are m is equal to 2 first one second braet 2 third one is min -2 these are Roots two roots are same third one is different now write in complimentary function is two roots are same means C1 + C2 X and two roots are same e to the power 2x plus third one is C3 into e - 2x so required compl function next example solve D 4 y by DX 4us M 4 y = to0 find we need a complementary function now solution the given equation is D 4 - M 4 into y = 0 d stand d by DX now it's auxiliary equation its auxiliary equation is no need for take D equal m so it get zero the auxilary equation means rate D 4 - M 4 = 0 now see this one we can write in this form d ² - m² and D ² + m² = 0 means uh a squ - b square formula a square square - b square square you get like this again a square - b square formula and this one is D- m M into d uh sorry D + m a b into a + b and this is also same uh D minus m i and D + m i equal to Z so how you get third and fourth bracket now c d ² is equal to M uh D ² + m² = 0 or d s = - m² or D = plus or- < TK of - m² but I is there so I means squ of minus1 okay now we get third and fourth bracket now roots are therefore the roots are D is equal to first one is plus M second one is minus M third one is a + I and fourth one is min - m i plus m i- m i now WR in a complimentary function is solution compliment function is the roots are different first and second C1 e to MX first root second one C2 e to Min - MX and third and fourth are a complex Roots Alpha is not there Alpha is zero plus Alpha is zero uh plus or minus m i uh these two third and fourth roots are plus or minus m r that means is Alpha is not there C3 cos mx + C4 sin MX this is required complementary function and next example solve D 4 + M 4 into y = 0 now see the previous example D 4us M 4 and this one is D 4 + M 4 only sign is different now see the problem how it work given equation [Music] is so D 4 + M 4 into y = 0 now o equation is auxiliary equation is D 4 + M 4 = 0 we not get factorization or take quadratic equation is not possible now in this example this type of the example adding a subtraction adding is + 2 m² D ² + M 4 and subtract 2 m² D ² = 0 take suitable equations okay now see this only first three terms and this is a d ² + m² S A + B S and - 2 m² D ² this is as it is equal to Zer now a squ - b square formula you apply it A- B into a + b now using that formula is a s + B squ a² + uh m² + < tk2 m d first one a + b and a minus b d ² + m² - < tk2 MD equal to0 Now using this these two bracket using a quadratic equation D = root 2 m + orus otk of b s - 4 a c 2 m² - 4 m² / 2 a 2 first one second one is D = to - B- -+ < tk2 M plus or- of b s - 4 a c 2 2 m² - 4 m² / 2 a and this one is I simplify - M by < tk2 + m i by < tk2 and this one is M by < tk2 plus orus m i by < tk2 so generally we taking quadratic equation a x² + BX + C = 0 and x equal and this one is D ² take D is equal to the generally qu equation x = - b + orus otk of b s - 4 a c/ 2 a this example is take D is equal to uh these two are a complex Roots Alpha plus orus I beta Alpha plus or- I beta first one is Alpha = minus sign second one is Alpha = plus sign now complementary function is the first one e to alpha x Alpha is - M by < tk2 into X into C1 cos beta this is beta part MX by < tk2 + C2 sin MX by < tk2 first one is what plus second same Alpha plus orus I beta using the compliment function Alpha is a m by < tk2 E power M by < tk2 into X C3 this is beta part COS of MX by < tk2 + C4 sin into MX by < tk2 now this is a required complementary function