Find the general solution of the given differential equation. y” − 2y' − 3y =

Find the general solution of the given differential equation. $$\displaystyle{y}”−{2}{y}'−{3}{y}={3}{e}^{{{2}{t}}}$$

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Jozlyn
First find solution of homogenenous problem.
$$\displaystyle{r}^{{2}}-{2}{r}-{3}={0}\to{r}_{{{1},{2}}}=-{1},{3}\to$$ using quadratic formula
Homogeneous solution:
$$\displaystyle{y}_{{c}}={c}_{{1}}{e}^{{-{t}}}+{c}_{{2}}{e}^{{{3}{t}}}$$
Let $$\displaystyle{Y}={A}{e}^{{{2}{t}}}$$ because $$\displaystyle{g{{\left({t}\right)}}}={3}{e}^{{{2}{t}}}$$
Plug Y into starting equation to find particular solution:
$$\displaystyle{\left({A}{e}^{{{2}{t}}}\right)}{''}-{2}{\left({A}{e}^{{{2}{t}}}\right)}'-{3}{A}{e}^{{{2}{t}}}={3}{e}^{{{2}{t}}}$$
$$\displaystyle{4}{A}{e}^{{{2}{t}}}-{4}{A}{e}^{{{2}{t}}}-{3}{A}{e}^{{{2}{t}}}={3}{e}^{{{2}{t}}}$$
$$\displaystyle-{3}{A}{e}^{{{2}{t}}}={3}{e}^{{{2}{t}}}$$
$$\displaystyle{A}=-{1}$$
$$\displaystyle{Y}=-{e}^{{{2}{t}}}$$
Solution: $$\displaystyle{y}={y}_{{c}}={c}_{{1}}{e}^{{-{t}}}+{c}_{{2}}{e}^{{{3}{t}}}-{e}^{{{2}{t}}}$$
Result: $$\displaystyle{y}={y}_{{c}}={c}_{{1}}{e}^{{-{t}}}+{c}_{{2}}{e}^{{{3}{t}}}-{e}^{{{2}{t}}}$$