ODE y''+y=4*x*sin x. Guessing particular solution?

[Acko93]

I can't realise how to guess particular solution for this eq. Can someone explain? I tried Axsinx, A(x^2)sinx, Axsinx + Bxcosx. Also tried transforming to complex but i just cant do it. Thanks.

y''+y=4*x*sin x

 

[tkhunny]

1) What is your complementary solution?
2) What do you think was wrong with $\displaystyle Ax\sin(x) + Bx\cos(x)$?

Please provide more detail into what you have done and what it looked like and where you think it failed.

 

[Acko93]

1) Complementary solution is c1sinx+c2cosx.
2) When I put back in yp'' and yp, terms with xsinx get nullified and i am left with sinx and cosx. So i get for A and B that they are 0.

 

[Acko93]

It is A(x^2)sinx + Bxsinx + C(x^2)cosx. I thought that i should multiply with x only if terms are same but i should even if 1 of them already has x with 1 more exponent.

 

[tkhunny]

It is A(x^2)sinx + Bxsinx + C(x^2)cosx. I thought that i should multiply with x only if terms are same but i should even if 1 of them already has x with 1 more exponent.

You have the right idea. Don't be discouraged. Keep going. You'll need and $\displaystyle x^{3}$ for this one. $\displaystyle x^{2}$ doesn't quite do it.

 

[HallsofIvy]

When you have a power of x, say power n, you will typically to try a polynomial of degree n including lower power terms.

Use (Ax+ B)cos(x)+ (Cx+ D)sin(x).
Then y'= A cos(x)- (Ax+ B) sin(x)+ C sin(x)+ (Cx+ D) cos(x)= (Cx+ A+ D)cos(x)- (Ax+ B- C)sin(x)

and y''= -2A sin(x)- (Ax+ B) cos(x)+ 2C cos(x)- (Cx+ D) sin(x)= (Cx+ D- 2A)sin(x)- (Ax+ B- 2C)cos(x)

So y''+ 4y= (Cx- Ax+ D- 2A- B+ C)sin(x)+ (Cx- Ax+ A+ D- B+ 2C)cos(x)

We want that to be equal to 4x sin(x). We want C- A= 4 , D- 2A- B+ C= 0, C- A= 0, A+ D- B+ 2C= 0, four equations to solve for A, B, C, and D.