# Solve the following differential equations. y" - 4y = 0

Solve the following differential equations.
$$\displaystyle{y}\text{}-{4}{y}={0}$$

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Linda Birchfield
Given: $$\displaystyle{y}\text{}{4}{y}={0}$$
$$\displaystyle\rightarrow{\frac{{{d}^{{{2}}}{y}}}{{{\left.{d}{x}\right.}^{{{2}}}}}}-{4}{y}={0}$$
Let $$\displaystyle{\frac{{{d}^{{{2}}}}}{{{\left.{d}{x}\right.}^{{{2}}}}}}={0}^{{{2}}}=$$ differential operator
$$\displaystyle{D}^{{{2}}}{y}-{4}{y}={0}$$
$$\displaystyle{\left({D}^{{{2}}}-{4}\right)}{y}={0}$$
For complementary sol put $$\displaystyle{D}^{{{2}}}-{4}={0}$$
$$\displaystyle{D}^{{{2}}}-{4}={0}$$
$$\displaystyle{D}^{{{2}}}={4}$$
$$\displaystyle{D}=\pm{2}$$
$$\displaystyle{m}_{{{1}}}={2}\ \text{and}\ {m}_{{{2}}}=-{2}$$
$$\displaystyle{y}={c}_{{{1}}}{e}^{{{m}{x}}}+{c}_{{{2}}}{e}^{{{m}{2}{x}}}$$
$$\displaystyle{c}_{{{1}}}\ \text{and}\ {c}_{{{2}}}=\text{constants}$$
Answer: $$\displaystyle{y}={c}_{{{1}}}{e}^{{{2}{x}}}+{c}_{{{2}}}{e}^{{-{2}{x}}}$$
###### Not exactly what you’re looking for?
Chanell Sanborn
$$\displaystyle{m}^{{{2}}}-{4}={0},{m}{1}={2},{m}{2}=-{2}$$.
$$\displaystyle{y}={C}{1}\cdot{e}^{{{2}{x}}}+{C}{2}\cdot{e}^{{-{2}{x}}},{\quad\text{or}\quad}\ {y}={C}{1}\cdot{c}{h}{\left({2}{x}\right)}+{C}{2}\cdot{s}{h}{\left({2}{x}\right)}$$