Particular Solution

Consider the nonhomogeneous differential equation $$${{y}}^{{{\left({n}\right)}}}+{a}_{{{n}-{1}}}{\left({x}\right)}{{y}}^{{{\left({n}-{1}\right)}}}+\ldots+{a}_{{1}}{\left({x}\right)}{y}'+{a}_{{0}}{\left({x}\right)}{y}={g{{\left({x}\right)}}}$$$.

Recall from the section about linear independence that the solution of the corresponding homogeneous equation $$${{y}}^{{{\left({n}\right)}}}+{a}_{{{n}-{1}}}{\left({x}\right)}{{y}}^{{{\left({n}-{1}\right)}}}+\ldots+{a}_{{1}}{\left({x}\right)}{y}'+{a}_{{0}}{\left({x}\right)}{y}={0}$$$ is given as $$${y}_{{h}}={c}_{{1}}{y}_{{1}}{\left({x}\right)}+{c}_{{2}}{y}_{{2}}{\left({x}\right)}+\ldots+{c}_{{n}}{y}_{{n}}{\left({x}\right)}$$$.

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 $$${y}={y}_{{h}}+{y}_{{p}}$$$ into the differential equation:

$$${{\left({y}_{{h}}+{y}_{{p}}\right)}}^{{{\left({n}\right)}}}+{a}_{{{n}-{1}}}{\left({x}\right)}{{\left({y}_{{h}}+{y}_{{p}}\right)}}^{{{\left({n}-{1}\right)}}}+\ldots+{a}_{{1}}{\left({x}\right)}{\left({y}_{{h}}+{y}_{{p}}\right)}'+{a}_{{0}}{\left({x}\right)}{y}=$$$

$$$={\left({{y}_{{h}}^{{{\left({n}\right)}}}}+{a}_{{{n}-{1}}}{\left({x}\right)}{{y}_{{h}}^{{{\left({n}-{1}\right)}}}}+\ldots+{a}_{{1}}{\left({x}\right)}{y}_{{h}}'+{a}_{{0}}{\left({x}\right)}{y}_{{h}}\right)}+{\left({{y}_{{p}}^{{{\left({n}\right)}}}}+{a}_{{{n}-{1}}}{\left({x}\right)}{{y}_{{p}}^{{{\left({n}-{1}\right)}}}}+\ldots+{a}_{{1}}{\left({x}\right)}{y}_{{p}}'+{a}_{{0}}{\left({x}\right)}{y}_{{p}}\right)}=$$$

$$$={0}+{g{{\left({x}\right)}}}={g{{\left({x}\right)}}}$$$

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 $$${y}''+{y}'-{2}{y}={0}$$$ is $$${y}_{{h}}={c}_{{1}}{{e}}^{{-{2}{x}}}+{c}_{{2}}{{e}}^{{{x}}}$$$. For the nonhomogeneous equation $$${y}''+{y}'-{2}{y}={2}{\left({1}+{x}-{{x}}^{{2}}\right)}$$$, the particular solution is $$${y}_{{p}}={{x}}^{{2}}$$$. So, the general solution of the nonhomogeneous differential equation is $$${y}''+{y}'-{2}{y}={2}{\left({1}+{x}-{{x}}^{{2}}\right)}$$$ is $$${y}={y}_{{h}}+{y}_{{p}}={c}_{{1}}{{e}}^{{-{2}{x}}}+{c}_{{2}}{{e}}^{{x}}+{{x}}^{{2}}$$$.