Particular Solution

Consider the nonhomogeneous differential equation `y^((n))+a_(n-1)(x)y^((n-1))+...+a_1(x)y'+a_0(x)y=g(x)` .

Recall from section about linear independence that solution of the corresponding homogeneous equation `y^((n))+a_(n-1)(x)y^((n-1))+...+a_1(x)y'+a_0(x)y=0` is given as `y_h=c_1y_1(x)+c_2y_2(x)+...+c_ny_n(x)` .

Then if `y_p` is solution of nonhomogeneous equation then general solution of nonhomogeneous equation is `y=y_h+y_p` .

Proof.

Plug in `y=y_h+y_p` into differential equation:

`(y_h+y_p)^((n))+a_(n-1)(x)(y_h+y_p)^((n-1))+...+a_1(x)(y_h+y_p)'+a_0(x)y=`

`=(y_h^((n))+a_(n-1)(x)y_h^((n-1))+...+a_1(x)y_h'+a_0(x)y_h)+(y_p^((n))+a_(n-1)(x)y_p^((n-1))+...+a_1(x)y_p'+a_0(x)y_p)=`

`=0+g(x)=g(x)`

So, indeed `y=y_p+y_h` is solution of nonhomogeneous differential equation.

Example. Solution of homogeneous equation `y''+y'-2y=0` is `y_h=c_1e^(-2x)+c_2e^(x)` .

For the nonhomogeneous equation `y''+y'-2y=2(1+x-x^2)` particular solution is `y_p=x^2` .

So, general solution of nonhomogeneous differential equation `y''+y'-2y=2(1+x-x^2)` is `y=y_h+y_p=c_1e^(-2x)+c_2e^x+x^2` .