# Thread: Solve y ′′ − 3 y ′ − 4 y = −3 x+5

1. ## Solve y ′′ − 3 y ′ − 4 y = −3 x+5

Solve the differential equation
y ′′ − 3 y ′ − 4 y = −3 x+5

I've worked out the independent solution which is
r^2 - 3r - 4 = 0

(r + 1) (r - 4)

{e^(-1*x), e^(4*x)}
(as a set of function)

Now my question is how do I find the solution of the original d.e. is of form
y(x) = u1v1 + u2v2

And the general solution?

2. Originally Posted by lemonlime
Solve the differential equation
y ′′ − 3 y ′ − 4 y = −3 x+5
This is a linear constant coefficient ordinary differential equation. The general solution is of the form:

$\displaystyle y(x)=G(x)+p(x)$

where $\displaystyle G(x)$ is the general solution of the homogeneous equation:

$\displaystyle y''-3y'-4y=0$

(you seem to know how to do this) and $\displaystyle p(x)$ is any particular solution of the full equation. To find a particular solution use a trial solution of the for $\displaystyle y(x)=ax+b$, plug this into the original equation and find what $\displaystyle a$ and $\displaystyle b$ need to be for the equation to be satisfied.

CB