Diff Eq- Nonhomogeneous Equations

[Totalderiv]

Homework Statement

Find a particular solution of the given equation.
$$y^''' + 4y^' = 3x-1$$

Homework Equations

$$r^3 + 4r = 0$$
$$r = 0, r = 2i, r = -2i$$

The Attempt at a Solution

$$y(x) = Ax-B$$
$$y^'(x) = A$$
$$y^''(x) = 0$$
$$y^'''(x) = 0$$


The answer is:
$$y(x)=(3/8)x^2 - (1/4)x$$
But I'm not sure how they came to this, please help!

 

[LCKurtz]

r=0 gives a constant as a solution of the homogeneous equation. So instead of trying Ax+B you must multiply by x and try $y_p=Ax^2+Bx$.

 

[LCKurtz]

Thanks! I have another question though,

$$4y^'' + 4y^' + y = 3xe^x$$

How do I start this?

The same way you started the other one. You find the complementary solution and then use Undetermined Coefficients for the particular solution. Surely your text discusses the method of Undetermined Coefficients, doesn't it?