Carole Yarbrough

2021-12-26

Find that solution of ${y}^{\prime }=2\left(2x-y\right)$ which passes through the point (0, 1).

Bukvald5z

Given ${y}^{\prime }=2\left(2x-y\right)\left(0,1\right)$
$\frac{dy}{dx}=2\left(2x-y\right)$
$\frac{dy}{dx}+2y=4x$
If $={e}^{\int pdx}={e}^{\int 2dx}={e}^{2x}$
$y\left(IF\right)=\int \left(4x\right)\left(IF\right)dx+c$
$y\left({e}^{2x}\right)=4\int x{e}^{2x}dx+c$
$y\left({e}^{2x}\right)=4\left[x\int {e}^{2x}dx-\int \left[\frac{d}{dx}\left(x\right)\int {e}^{2x}dx\right]dx+c$
$y\left({e}^{2x}\right)=4\left[x\frac{{e}^{2x}}{2}-\int \frac{{e}^{2x}}{2}dx\right]+c$
$y\left({e}^{2x}\right)=4\left[\frac{x{e}^{2x}}{2}-\frac{{e}^{2x}}{4}\right]+c$
$y\left({e}^{2x}\right)=\frac{4}{4}\left[2x{e}^{2x}-{e}^{2x}\right]+c$
$y\left({e}^{2x}\right)=\left[2x{e}^{2x}-{e}^{2x}\right]+c$ (a)
$y\left(0\right)=1$ from (a)
$1\left({e}^{0}\right)=\left[2.0{e}^{0}-{e}^{0}\right]+c$
$1=\left[0-1\right]+c$
$2=c$
From (a)
$y\left({e}^{2x}\right)=\left[2x{e}^{2x}-{e}^{2x}\right]+2$

Durst37

Given: The differential equation ${y}^{\prime }=2\left(2x-y\right)$ and the point P(0,1).
Solution: ${y}^{\prime }=2\left(2x-y\right)$
$=4x-2y$
${y}^{\prime }+2y=4x$
$\frac{dy}{dx}+2y=4x$
The above differential equation is in the form of first order linear differential equation.
$\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$
Here
Integrating factor $\left(IF\right)={e}^{\int P\left(x\right)dx}$
$={e}^{\int 2dx}$
$={e}^{2x}$
The general solution is
$y×IF=\int \left(IF×Q\left(x\right)\right)dx$
$y×{e}^{2x}=\int \left({e}^{2x}×4x\right)dx$
$y×{e}^{2x}=2{e}^{2x}x-{e}^{2x}+C$
$y=\frac{2{e}^{2x}x-{e}^{2x}+C}{{e}^{2x}}$ passes through the point P(0,1)
$1=\frac{2{e}^{2\left(0\right)}\left(0\right)-{e}^{2\left(0\right)+C}}{{e}^{2\left(0\right)}}$

$1=\frac{0-1+C}{1}$
$1=-1+C$
$C=2$
The solution of the differential equation ${y}^{\prime }=2\left(2x-y\right)$ is
$⇒y\left(x\right)=\frac{2{e}^{2x}x-{e}^{2x}+2}{{e}^{2x}}$
$⇒y\left(x\right)=2x-1+2{e}^{-2x}$

user_27qwe