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.

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.