Question

# Solve the differential equation. (x and y are variables) (D^2+4D)y=1+2x+3x^2

Multivariable functions
Solve the differential equation. (x and y are variables)
$$\displaystyle{\left({D}^{{2}}+{4}{D}\right)}{y}={1}+{2}{x}+{3}{x}^{{2}}$$

2021-09-15

Given differential equation is $$\displaystyle{\left({D}^{{2}}+{4}{D}\right)}{y}={1}+{2}{x}+{3}{x}^{{2}}$$ - (1)
$$\displaystyle{y}\text{}{4}{y}'={1}+{2}{x}+{3}{x}^{{2}}$$
The auxillary equation is
$$\displaystyle{m}^{{2}}+{4}{m}={0}$$
Step 2
$$\displaystyle{m}^{{2}}+{4}{m}={0}$$
$$\displaystyle{m}{\left({m}+{4}\right)}={0}$$
$$\displaystyle{m}={0}.{m}=-{4}$$
$$\displaystyle{y}_{{c}}={c}_{{1}}{e}^{{{0}{x}}}+{c}_{{2}}{e}^{{-{4}{x}}}$$
$$\displaystyle\Rightarrow{y}_{{c}}={c}_{{1}}+{c}_{{2}}{e}^{{-{4}{x}}}$$
Then $$\displaystyle{y}_{{p}}={A}{x}^{{2}}+{B}{x}$$ by using method of undetermined coefficient
$$\displaystyle{y}_{{p}}'={2}{A}{x}+{B}$$
$$\displaystyle{y}_{{p}}\text{}{2}{A}$$
Then $$\displaystyle{y}_{{p}}\text{}{4}{y}_{{p}}'={1}+{2}{x}+{3}{x}^{{2}}$$
$$\displaystyle{2}{A}+{2}{A}{x}+{B}={1}+{2}{x}+{B}{x}^{{2}}$$
$$\displaystyle{\left({2}{A}+{B}\right)}+{2}{A}{x}={3}{x}^{{2}}+{2}{x}+{1}$$
Comparing the coefficients
Step 3

$$\displaystyle{2}{A}={2}.\Rightarrow{A}={1}$$
$$\displaystyle{2}{A}+{B}={1}\Rightarrow{2}{\left({1}\right)}+{B}={1}$$
$$\displaystyle{B}={1}-{2}$$
$$\displaystyle{B}=-{1}$$
Then $$\displaystyle{y}_{{p}}={x}^{{2}}-{x}$$
The general general solution is
$$\displaystyle{y}={y}_{{c}}+{y}_{{p}}$$
$$\displaystyle{y}+{c}_{{1}}+{c}_{{2}}{e}^{{-{4}{x}}}+{x}^{{2}}-{x}$$ Isregmined solution