# One solution of the differential equation y" – 4y = 0 is y = e^{2x} Find a second linearly independent solution using reduction of order.

e1s2kat26 2021-01-28 Answered

One solution of the differential equation $y"–4y=0$ is $y={e}^{2x}$Find a second linearly independent solution using reduction of order.

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## Expert Answer

krolaniaN
Answered 2021-01-29 Author has 86 answers
Given differential equation is ${y}^{″}-4y=0$ (1)
Given that First solution is ${y}_{1}={e}^{2x}$
let general solution $y=v\cdot {y}_{1}$
substitute ${y}_{1}={e}^{2x}$
$y=v{e}^{2x}$
Differentiate $y=v{e}^{2x}$ with respect to x
${y}^{\prime }=\frac{d}{dx}\left(v{e}^{2x}\right)$
$=\frac{vd}{dx}{e}^{2x}+{e}^{2x}\frac{d}{dx}\left(v\right)$
$=v\cdot 2{e}^{2x}+{e}^{2x}\cdot {v}^{\prime }$
$=2v{e}^{2x}+{e}^{2x}{v}^{\prime }$
hence, ${y}^{\prime }=2v{e}^{2x}+{e}^{2x}v,$
Again differentiate ${y}^{\prime }=2v{e}^{2x}+{e}^{2x}{v}^{\prime }$ with respect to x
${y}^{″}=\frac{d}{dx\left(2v{e}^{2x}+{e}^{2x}{v}^{\prime }\right)}$
$=2\frac{d}{dx}\left(v{e}^{2x}\right)+\frac{d}{dx}\left({e}^{2x}{v}^{\prime }\right)$
$=4v{e}^{2x}+2{e}^{2x}{v}^{\prime }+2{e}^{2x}{v}^{\prime }+{e}^{2x}{v}^{″}$
$=4v{e}^{2x}+4{e}^{2x}{v}^{\prime }+{e}^{2x}{v}^{″}$
Hence, ${y}^{″}=4v{e}^{2x}+4{e}^{2x}{v}^{\prime }+{e}^{2x}{v}^{″}$
Now, substitute ${y}^{″}=4v{e}^{2x}+4{e}^{2x}{v}^{\prime }+{e}^{2x}{v}^{″}$ and $y=v{e}^{2x}$ in equation (1) and simlify it
$4v{e}^{2x}+4{e}^{2x}{v}^{\prime }+{e}^{2x}{v}^{″}-4v{e}^{2x}=0$
$4{e}^{2x}{v}^{\prime }+{e}^{2x}{v}^{″}=0$ (2)
Let $w={v}^{\prime }$
$⇒{w}^{\prime }={v}^{″}$
Substitute these value in equation (2)
$4{e}^{2x}w+{e}^{2x}{w}^{\prime }=0$
${e}^{2x}\left(4w+{w}^{\prime }\right)=0$ (Take common as ${e}^{2x}$)
Divide both side be ${e}^{2x}$ and simplify it
$\frac{{e}^{2x}\left(4w+{w}^{\prime }\right)}{{e}^{2x}}=\frac{0}{{e}^{2x}}$
$4w+{w}^{\prime }=0$
Substract 4w from both sides and simplify it
$4w-4w+{w}^{\prime }=-4w$
${w}^{\prime }=-4w$
$\frac{dw}{dx}=-4w$
$dww=-4dx$
Now integrate it
$\int \frac{dw}{w}=-4\int dx+c$
$\mathrm{ln}w=-4x+c$
$w={e}^{-4x+c}$
$w={e}^{-4x}\cdot {e}^{c}$
Further simplify it
$w={c}_{1}{e}^{-4x}$ (Since

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