Question

Can I please have help? Urgent question

Solve the differential equation by Undetermined Coefficients. y''+y=2xsinx

I need a detailed walkthrough, cant find the particular solution.

Answer 1

Okay, I'll assume you have the right homogeneous solution. As your guess, try \(y_p=A\sin x+B\cos x+Cx\sin x+Dx\cos x\).

Answer 2

is that the correct form? I thought it was (Ax^2+Bx)sinx + (Cx^2+Dx)cosx

Answer 3

Let's say, as an example, that you were given \(2t+3\). Would your guess be \(At^2+Bt\), or \(At+B\)?

Answer 4

\[At+B\]

Answer 5

Right. It's for the same reason that you would pick the guess solution I suggested. You could also try \((Ax^2+Bx+C)\sin x+(\cdots)\cos x\), but you'll likely find that the quadratic constants are 0.

Answer 6

So what would Yp' and Yp'' look like. (using the correct form of Yp)

Answer 7

\[Yp=A \cos x+Bsinx+Cxcosx+Dxsinx\]
\[Yp'=-Asinx+Bcosx+Ccosx-Cxsinx+Dsinx-Dxsinx\]
\[Yp''=-Acosx-Bsinx-Csinx-Csinx-Cxcosx+Dcosx-Dsinx-Dxcosx\]

Answer 8

The last term in \(y_p'\) isn't right, and I think an error follows from that in \(y_p''\).
\[\begin{align*}
y_p&=A\sin x+B\cos x+Cx\sin x+Dx\cos x\\
y_p'&=A\cos x-B\sin x+C\sin x+Cx\cos x+D\cos x-Dx\sin x\\
&=(A+D)\cos x+(C-B)\sin x+Cx\cos x-Dx\sin x\\
y_p''&=-(A+D)\sin x+(C-B)\cos x+C\cos x-Cx \sin x-D\sin x-Dx\cos x\\
&=-(A+2D)\sin x+(2C-B)\cos x-Cx\sin x-Dx\cos x
\end{align*}\]

Answer 9

I dont get how Yp is \[\frac{ 1 }{ 2 }x^2\cos x+\frac{ 1 }{ 2 }x \sin x\]

Answer 10

Hmm, looks like I might have to brush up on my guess solutions. Just to be safe, I'd use
\[y_p=(Ax^2+Bx+C)\sin x+(Dx^2+Ex+F)\cos x\]