Second Order Nonhomogeneous Differential Equation (Method of Undetermined Coefficients) Find the

Kaiya Hardin 2022-04-30 Answered
Second Order Nonhomogeneous Differential Equation (Method of Undetermined Coefficients)
Find the general solution of the following Differential equation y2y+10y=excos(3x). We know that the general solution for 2nd order Nonhomogeneous differential equations is the sum of yp+yc where yc is the general solution of the homogeneous equation and yp the solution of the nonhomogeneous. Therefore yc=ex(c1cos(3x)+c2cos(3x)). Now we have to find yp. I know in fact that yc=ex(c1cos(3x)+c2cos(3x)). Now we have to find yp. I know in fact that yp=e×sin(3x)6 but i do not know how to get there.
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Answers (2)

j3jell5
Answered 2022-05-01 Author has 17 answers
Step 1
since the RHS is a soln of the homogeneous eqn, we can try x multiplied by it. let yp be a linear combination of xexcos(3x) and xexsin(3x).
yp=axexcos(3x)+bxexsin(3x)
Step 2
Let E=D22D+10, where D is the derivative operator.
E[yp]=6aexsin(3x)+6bexcos(3x)
E[yp]=excos(3x)
equating coeffs, a=0 and 6b=1
yp=16xexsin(3x)
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Bethany Mills
Answered 2022-05-02 Author has 14 answers
Step 1
After a lot of trial and error i finnaly understand the strategy behind this type of differential equation. So here's is what i did in order to find the solution (I will provide as much information as i can) In order to calculate a nonhomogeneous differntial equation we must first find the general solution of the homogeneous. So y2y+10y=o
(1) using the Auxiliary equation we can easily find that the general solution is yh=ex(C1cos(3x)+C2sin(3x). Now as cineel mentioned above we can try yh multiplied by x. Thus yp=Axexcos(3x)+Bxexsin(3x). Now we have to differentiate yp two times.
So yp=Axexcos(3x)+Bxexsin(3x) (1)
yp=A[excos(3x)+xexcos(3x)xex3sin(3x)]+B[exsin(3x)+xexsin(3x)+3xexcos(3x) (2)
yp=A[8xexcos(3x)6e×sin(3x)+2excos(3x)3exsin(3x)]+B[8xexsin(3x)+6e×cos(3x)+2exsin(3x)+6excos(3x)] (3)
Step 2
Therefore substituting (1)(2)(3) in our original differential equation we end up with 3Aexsin(3x)+6Bexcos(3x)=excos(3x)
So A=0 and B=16
Thus (1) Now becomes yp=16xexsin(3x)
So yp+yh=ex[C1cos(3x)+C2sin(3x)]+16e×sin(3x)
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