How to solve this 2nd order linear ODE

[Another1]

\(\displaystyle y''(x)+y'(x)+F(x)=0\)

Pleas me a idea

 

[topsquark]

\(\displaystyle y''(x)+y'(x)+F(x)=0\)

Pleas me a idea

Let y'(x) = v(x). Then your equation becomes v'(x) + v(x) = -F(x)

Now you have a first degree linear ordinary differential equation. You can't write down a final answer, but there is a method by which you can write it out in terms of F(x)...

-Dan

 

[HallsofIvy]

Since this has been sitting for a while now: The equation v'+ v= -F(x) has associated homogeneous equation v'+ v= 0 which has characteristic equation r+ 1= 0. r= -1 so the general solution to the associated homogeneous equation is $v= Ce^{-t}$.
(You could also see that by writing the equation as $v'= \frac{dv}{dt}=-v$ so that $\frac{dv}{v}= -dt$ and integrating: $ln(v)= -t+ C'$ so $v= e^{-t+ C'}= Ce^{-t}$ with $C= e^{C'}$.)

To find a solution to the entire equation, look for a solution of the form $v(t)= u(t)e^{-t}$ where u is a function to be found. $v'= u'e^{-t}- ue^{-t}$ so $v'+ v= u'e^{-t}- ue^{-t}+ ue^{-t}= u'e^{-t}= -F$. Then $u'= =Fe^{t}$ and $u= -\int Fe^{t}dt$.

The general solution to the entire equation is the general solution to the associated homogeneous equation plus that solution to the entire equation:
$v= y'= Ce^{-t}- \left(\int_0^t F(s)e^sds\right)e^{-t}$

Now y is the integral of that.