Question

Consider the nonhomogeneous differential

equationy"-2y'-3y=2(e^(3x)).
Find a paricular solution to the non homogeneous equation.
also write the general solutiong to the
nonhomogeneousequation.

Answer 1

`c_1e^(3x)+c_2e^(-x)+(1)/(2) xe^(3x)`

Explanation:

For homogeneous solution:

`y''-2y'-3=0\\r^2-2r-3=0\\(r-3)(r+1)=0`

So since roots are r = 3 and r = -1,
`y_h=c_1e^(3x)+c_2e^(-x)`

Since we are given
`2e^(3x)`, we will use undetermined coefficients. However, here the trick is we have
`c_1e^(3x)` in homogeneous solution. So in particular solution, as undetermined coefficients, we will use
`Axe^(3x)` instead of
`Ae^(3x)`.

Hence,

`Y = Axe^(3x)\\Y' = Ae^(3x) + 3Axe^(3x)\\Y'' = 3Ae^(3x) + 3Ae^(3x) + 9Axe^(3x) = 6Ae^(3x) + 9Axe^(3x)`

So,

`6Ae^(3x)+9Axe^(3x)-2Ae^(3x)-6Axe^(3x)-3Axe^(3x)=2e^(3x)\\4Ae^(3x)=2e^(3x)\\A=(1)/(2)`

Hence,
`y_p = (1)/(2) xe^(3x)`

General solution is:

`y=y_h+y_p=c_1e^(3x)+c_2e^(-x)+(1)/(2) xe^(3x)`