Question

# Consider the following differential equation. 2y''+y'-y=0 For what values of r does the function y = e^{rx} satisfy the equation?

Equations
Consider the following differential equation.
2y''+y'-y=0
For what values of r does the function $$\displaystyle{y}={e}^{{{r}{x}}}$$ satisfy the equation?

2021-05-18
Step 1
Given,
Consider the following differential equation 2y''+y'-y=0 and $$\displaystyle{y}={e}^{{{r}{x}}}$$.
Step 2
Now,
$$\displaystyle{y}={e}^{{{r}{x}}}$$
Differentiating on both sides, we get
$$\displaystyle\Rightarrow{y}'={\frac{{{d}{\left({e}^{{{r}{x}}}\right)}}}{{{\left.{d}{x}\right.}}}}$$
$$\displaystyle\Rightarrow{y}'={e}^{{{r}{x}}}{\frac{{{d}{\left({r}{x}\right)}}}{{{\left.{d}{x}\right.}}}}$$
$$\displaystyle\Rightarrow{y}'={e}^{{{r}{x}}}{r}$$
Again, differentiating on both sides, we get
$$\displaystyle\Rightarrow{y}{''}={r}{\frac{{{d}{\left({e}^{{{r}{x}}}\right)}}}{{{\left.{d}{x}\right.}}}}$$
$$\displaystyle\Rightarrow{y}{''}={r}^{{{2}}}{e}^{{{r}{x}}}$$
$$\displaystyle\therefore{2}{y}{''}+{y}'-{y}={0}$$
$$\displaystyle\Rightarrow{2}{r}^{{{2}}}{e}^{{{r}{x}}}+{r}{e}^{{{r}{x}}}-{e}^{{{r}{x}}}={0}$$
$$\displaystyle\Rightarrow{2}{r}^{{{2}}}+{r}-{1}={0}$$
$$\displaystyle\Rightarrow{2}{r}^{{{2}}}+{2}{r}-{r}-{1}={0}$$
$$\displaystyle\Rightarrow{2}{r}{\left({r}+{1}\right)}-{1}{\left({r}+{1}\right)}={0}$$
$$\displaystyle\Rightarrow{\left({2}{r}-{1}\right)}{\left({r}+{1}\right)}={0}$$
Either 2r-1=0 or r+1 = 0
$$\displaystyle{r}={\frac{{{1}}}{{{2}}}}{\quad\text{or}\quad}{r}=-{1}$$
$$\displaystyle\therefore$$ The value of r is $$\displaystyle{\frac{{{1}}}{{{2}}}}$$ and -1.