 # Solve differential equation xy'= 6y+12x^4 y^(2/3) Yasmin 2020-11-09 Answered
Solve differential equation$x{y}^{\prime }=6y+12{x}^{4}{y}^{\left(}2/3\right)$
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$dy/dx+P\left(x\right)y=Q\left(x\right){y}^{n}$
$n=0$ or $n=1$
Substituting $v={y}^{1}-n$ transforms the above equation into the linear equation
$\left(dv\right)/dx+\left(1-n\right)P\left(x\right)v=\left(1-n\right)Q\left(x\right)$
$\left(dv\right)/dx+P\left(x\right)y=Q\left(x\right)$
$y\left(x\right)={e}^{\left(\int P\left(x\right)dx\right)}\left[\int \left(Q\left(x\right){e}^{\left(\int P\left(x\right)dx\right)\right)}dx+C\right]$
The differential equation is,
$x{y}^{\prime }=6y+12{x}^{4}{y}^{\left(2/3\right)}$
The above equation can be written as follows
${y}^{\prime }-6/xy=12{x}^{3}{y}^{\left(2/3\right)}$
Substituting $v={y}^{1}-2/3$ or y $1/3$ transforms the above Bernoulli equation into the linear equation
$\left(dv\right)/dx-2/xv=4{x}^{3}$
$v\left(x\right)={e}^{\left(-\int \left(2/x\right)dx\right)}\left[\int \left(4{x}^{3}{e}^{\left(\int \left(-2/x\right)dx\right)\right)}dx+{C}^{\prime }\right]$
$={e}^{\left(2\mathrm{ln}x\right)}\left[\int \left(4{x}^{3}{x}^{-}2\mathrm{ln}x\right)dx+{C}^{\prime }\right]$
$={x}^{2}\left[4{x}^{2}/x+{C}^{\prime }\right]$
$v\left(x\right)=2{x}^{4}+C{x}^{2}$
$y1/3=2{x}^{4}+C{x}^{2}$
$y\left(x\right)=\left(2{x}^{4}+C{x}^{2}{\right)}^{3}$
where C is constant

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