# Thread: Inhomogeneous linear second-order ODE (constant coefficients)

1. ## Inhomogeneous linear second-order ODE (constant coefficients)

y''(x) +3(y'(x))+2y=4e^(2x)+7

this is my ODE, it is inhomogeneous linear and second order with constant coefficients.

have i got this correct in saying the general solution is:

y=(1/3)*e^(2x)+7/2 ???

please could someone verify this quickly for me? im just curious as to whether or not i should be adding some unknown constants or other terms as well?

thanks

2. Originally Posted by situation
y''(x) +3(y'(x))+2y=4e^(2x)+7

this is my ODE, it is inhomogeneous linear and second order with constant coefficients.

have i got this correct in saying the general solution is:

y=(1/3)*e^(2x)+7/2 ???

please could someone verify this quickly for me? im just curious as to whether or not i should be adding some unknown constants or other terms as well?

thanks
Take a look at the characteristic equation for the homogeneous problem:
$m^2 + 3m + 2 = 0$

There are two solutions to this equation, meaning you have two terms in your homogeneous solution. Calling these solutions a and b, then the homogeneous solution will be
$y_h(x) = Ae^{ax} + Be^{bx}$

-Dan

3. Originally Posted by topsquark
Take a look at the characteristic equation for the homogeneous problem:
$m^2 + 3m + 2 = 0$

There are two solutions to this equation, meaning you have two terms in your homogeneous solution. Calling these solutions a and b, then the homogeneous solution will be
$y_h(x) = Ae^{ax} + Be^{bx}$

-Dan
thanks!

4. Originally Posted by situation
y''(x) +3(y'(x))+2y=4e^(2x)+7

this is my ODE, it is inhomogeneous linear and second order with constant coefficients.

have i got this correct in saying the general solution is:

y=(1/3)*e^(2x)+7/2 ???
You certainly would be wrong to say this the general solution! A "general" solution to a differential equation always contains undetermined constants such that all solutions can be got by taking different values for those solutions. Perhaps you meant to say "particular" solution. That's easy to check. If $y(x)= \frac{1}{3}e^{2x}+ \frac{7}{2}$, then $y'(x)= \frac{2}{3}e^{2x}$, and $y''= \frac{4}{3}e^{2x}$.

Putting those into the given differential equation gives
$y''+ 2y'+ 2y= \frac{4}{3}e^{2x}+ 2e^{2x}+ \frac{2}{3}e^2x+ 7= 4e^{2x}+ 7$
which does, in fact, satisfy the equation. Yes, that is a particular solution.

Now, the "general solution" to the entire equation can be written by adding that to the general solution to the associated homogeneous equation, $y''+ 3y'+ 2y= 0$.

please could someone verify this quickly for me? im just curious as to whether or not i should be adding some unknown constants or other terms as well?

thanks