Question

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

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asked 2021-05-16
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?

Answers (1)

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.
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