Haven

2021-05-16

Consider the following differential equation.
2y+y-y=0

Aubree Mcintyre

Step 1
Given,
Consider the following differential equation 2y''+y'-y=0 and $y={e}^{rx}$.
Step 2
Now,
$y={e}^{rx}$
Differentiating on both sides, we get
$⇒{y}^{\prime }=\frac{d\left({e}^{rx}\right)}{dx}$
$⇒{y}^{\prime }={e}^{rx}\frac{d\left(rx\right)}{dx}$
$⇒{y}^{\prime }={e}^{rx}r$
Again, differentiating on both sides, we get
$⇒y{}^{″}=r\frac{d\left({e}^{rx}\right)}{dx}$
$⇒y{}^{″}={r}^{2}{e}^{rx}$
$\therefore 2y{}^{″}+{y}^{\prime }-y=0$
$⇒2{r}^{2}{e}^{rx}+r{e}^{rx}-{e}^{rx}=0$
$⇒2{r}^{2}+r-1=0$
$⇒2{r}^{2}+2r-r-1=0$
$⇒2r\left(r+1\right)-1\left(r+1\right)=0$
$⇒\left(2r-1\right)\left(r+1\right)=0$
Either 2r-1=0 or r+1 = 0
$r=\frac{1}{2}\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}r=-1$
$\therefore$ The value of r is $\frac{1}{2}$ and -1.

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