Haven

2021-05-16

Consider the following differential equation.

2y+y-y=0

2y+y-y=0

Aubree Mcintyre

Skilled2021-05-18Added 73 answers

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

$\Rightarrow {y}^{\prime}=\frac{d\left({e}^{rx}\right)}{dx}$

$\Rightarrow {y}^{\prime}={e}^{rx}\frac{d\left(rx\right)}{dx}$

$\Rightarrow {y}^{\prime}={e}^{rx}r$

Again, differentiating on both sides, we get

$\Rightarrow y{}^{\u2033}=r\frac{d\left({e}^{rx}\right)}{dx}$

$\Rightarrow y{}^{\u2033}={r}^{2}{e}^{rx}$

$\therefore 2y{}^{\u2033}+{y}^{\prime}-y=0$

$\Rightarrow 2{r}^{2}{e}^{rx}+r{e}^{rx}-{e}^{rx}=0$

$\Rightarrow 2{r}^{2}+r-1=0$

$\Rightarrow 2{r}^{2}+2r-r-1=0$

$\Rightarrow 2r(r+1)-1(r+1)=0$

$\Rightarrow (2r-1)(r+1)=0$

Either 2r-1=0 or r+1 = 0

$r=\frac{1}{2}{\textstyle \phantom{\rule{1em}{0ex}}}\text{or}{\textstyle \phantom{\rule{1em}{0ex}}}r=-1$

$\therefore$ The value of r is $\frac{1}{2}$ and -1.

Given,

Consider the following differential equation 2y''+y'-y=0 and

Step 2

Now,

Differentiating on both sides, we get

Again, differentiating on both sides, we get

Either 2r-1=0 or r+1 = 0

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