# 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 the section about linear independence that the 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 the solution of the nonhomogeneous equation, the general solution of the nonhomogeneous equation is y=y_h+y_p.

Proof.

Plug in y=y_h+y_p into the 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 the solution of the nonhomogeneous differential equation.

Now, let's consider an example.

Example. The solution of the 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), the particular solution is y_p=x^2.

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