Thread: find the general solution ODE pls help

$\displaystyle d^2y/dx^2 - dy/dx - dy= 3e^-x + 10sinx -4x$

please check if my answer are correct

$\displaystyle y=c1e^-x + c2e^2x -xe^-x+ cosx +3sinx+2x$

Originally Posted by tak

$\displaystyle d^2y/dx^2 - dy/dx - dy= 3e^-x + 10sinx -4x$

please check if my answer are correct

$\displaystyle y=c1e^-x + c2e^2x -xe^-x+ cosx +3sinx+2x$

Hello tak,
You can substitute y back in the differential equation and see if it holds.This way you can be more independent(you can even do this in exams )

So try it

Does it mean I have to differentiate again?

Originally Posted by tak

$\displaystyle d^2y/dx^2 - dy/dx - dy= 3e^-x + 10sinx -4x$

please check if my answer are correct

$\displaystyle y=c1e^-x + c2e^2x -xe^-x+ cosx +3sinx+2x$

I think we may have a problem.

$\displaystyle \frac{d^2y}{dx}-\frac{dy}{dx}-y=3e^{-x}+10\sin(x)-4x$

solving the homogenious equation for the particular solution we get..

$\displaystyle \frac{d^2y}{dx}-\frac{dy}{dx}-y=0$

$\displaystyle m^2-m-1=0 \iff m^2-m+\frac{1}{4}=1+\frac{1}{4} \iff (x-\frac{1}{2})^2=\frac{5}{4} \iff x=\frac{1 \pm \sqrt{5}}{2}$

so

$\displaystyle y_c=c_1e^{\frac{1+\sqrt{5}}{2}x}+c_2e^{\frac{1-\sqrt{5}}{2}x}$

See what you can do from here.

$\displaystyle \frac{d^2y}{dx}-\frac{dy}{dx}-2y=3e^{-x}+10\sin(x)-4x
$
very sorry it should be 2y

Originally Posted by tak

$\displaystyle \frac{d^2y}{dx}-\frac{dy}{dx}-2y=3e^{-x}+10\sin(x)-4x
$
very sorry it should be 2y

then the equation is

$\displaystyle m^2-m-2=0 \iff (m-2)(m+1)$

so

$\displaystyle y_c=c_1e^{2x}+c_2e^{-x}$

Now we need to find the particular solution

since $\displaystyle e^{-x}$ in the complimentry solution
the particular solution will be of the form

$\displaystyle y_p=\underbrace{Ax^2+Bx+C}_{forThe-4x}+\underbrace{E\cos(x)+F\sin(x)}_{forThe10\sin(x )}+\underbrace{G(e^{-x})+H(xe^{-x})}_{forTheRepeted 3 e^{-x}}$

Hi

I Know that i need to multiply by x but do I need to keep the existing term for the last part.

how do you know that you need to keep the G(e^-x) term. I saw some website they don't add in and some does. Can I know th rule