 Daniell Phillips

2021-12-29

Find the solution of the following Differential Equations ${y}^{\prime }-4y=2x-4{x}^{2}$ censoratojk

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Find the solution of the following DE
${y}^{\prime }-4y=2x-4{x}^{2}$
Complete solution $=CF+PI$
CF=y'-4y=0
$\left(D-4\right)y=0$
$m-4=0$
$m=4$
$CF={C}_{1}{e}^{4x}$
$PI=\frac{a}{f\left(D\right)}⇒\frac{2x-4{x}^{2}}{D-4}$
$PI=\frac{1}{-4+D}\left(2x-4{x}^{2}\right)$
$=\frac{1}{-4\left(1-\frac{D}{4}\right)}\left(2x-4{x}^{2}\right)-4+D=-4\left(1-\frac{D}{4}\right)$
$=-\frac{1}{4}{\left(1-\frac{D}{4}\right)}^{-1}\left(2x-4{x}^{2}\right)$
${\left(1-D\right)}^{-1}=1+D+{D}^{2}+{D}^{3}+\dots$
${\left(1+D\right)}^{-1}=1-D+{D}^{2}-{D}^{3}+\dots$
$D\left(2x-4{x}^{2}\right)=\frac{d}{dx}\left(2x-4{x}^{2}\right)=2-8x$
${D}^{2}\left(2x-4{x}^{2}\right)=\frac{{d}^{2}}{{dx}^{2}}\left(2x-4{x}^{2}\right)=-8$
${D}^{3}\left(2x-4{x}^{2}\right)=\frac{{d}^{3}}{{dx}^{3}}\left(2x-4{x}^{2}\right)=0$
$PI=-\frac{1}{4}\left(1+\frac{D}{4}+{\left(\frac{D}{4}\right)}^{2}+{\left(\frac{D}{4}\right)}^{3}+\dots \right)\left(2x-4{x}^{2}\right)$ jean2098

Expert

${y}^{\prime }-4y=2x-4{x}^{2}$ (1)
Clearly, the given differential equation is of the form ${y}^{\prime }+py=a$
Thus, differential eq (1) is linear, where
Now, integrating factor, $I.F={e}^{\int pdx}={e}^{\int dx}$
$⇒I.F={e}^{-4x}$
The solution of given differential equation (1) is given by,
$y\left(IF\right)=\int q\left(IF\right)dx+c$, c is a constant
$⇒y.{e}^{-4x}=\int \left(2x-4{x}^{2}\right){e}^{-4x}dx+c$ (2)
Now $\int \left(2x-4{x}^{2}\right){e}^{-4x}dx=\int \frac{{e}^{u}u\left(u+2\right)}{16}du$
$⇒\int {\left(2x-4{x}^{2}\right)}_{edx}^{-4x}\int {e}^{u}u\left(u+2\right)du$ (3)
$\frac{1}{16}\int {e}^{u}u\left(u+2\right)du=\frac{1}{16}\left[\left(u+2\right)\left({e}^{u}u-{e}^{u}\right)-\int \left({e}^{u}u-{e}^{u}\right)du\right]$
$=\frac{1}{16}\left[\left(u+2\right)\left({e}^{u}u-{e}^{u}\right)-{e}^{u}u+2{e}^{u}\right]$
$=\frac{1}{16}\left[{u}^{2}{e}^{u}-u{e}^{u}+2u{e}^{u}-2{e}^{u}-u{e}^{u}+2{e}^{u}\right]$
$=\frac{1}{16}{u}^{2}{e}^{u}$
Hence, from eq (3), we get
$\int \left(2x-4{x}^{2}\right){e}^{-4x}x=\frac{1}{16}{u}^{2}{e}^{u}$
$u=-4x$, we get karton

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