Solve $x . \frac{d y}{d x} - y = 2 x^{2} y$ $x.dy/dx-y=2x^2y$ ?

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Solve $x . \frac{d y}{d x} - y = 2 x^{2} y$ $x.dy/dx-y=2x^2y$ ?

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Answer 1 · 2018-02-18T16:29:50

$y = x e^{[x^{2} + c]}$ $y=xe^[[x^2+c]]$

Explanation:

$x \frac{d y}{d x} = 2 x^{2} y + y$ $xdy/dx=2x^2y+y$ , separating the variables.....

$\frac{1}{y} \frac{d y}{d x} = [2 x^{2} + \frac{1}{x}]$ $1/ydy/dx=[2x^2+1/x]$ so $\frac{d y}{y} = 2 x^{2} d x + \frac{d x}{x}$ $dy/y=2x^2dx+dx/x$ , integrating...
$\int \frac{1}{y} d y = \int 2 x^{2} d x + \int \frac{1}{x} d x$ $int1/ydy=int2x^2dx+int1/xdx$ , we have $ln y = x^{2} + ln x + C$ $lny=x^2+lnx+C$

Which gives $ln [\frac{y}{x}] = x^{2} + C$ $ln[y/x]=x^2+C$ .

$e^{ln [\frac{y}{x}]} = e^{[x^{2} + c]}$ $e^[ln[y/x]]= e^[[x^2+c]]$ [ theory of logs]

i.e, $\frac{y}{x} = e^{[x^{2} + c]}$ $y/x= e^[[x^2+c]]$ and $s o, y = x e^{[x^{2} + c]}]$ $so, y=xe^[[x^2+c]]]$

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Solve $x . \frac{d y}{d x} - y = 2 x^{2} y$ $x.dy/dx-y=2x^2y$ ?

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Solve $x . \frac{d y}{d x} - y = 2 x^{2} y$ $x.dy/dx-y=2x^2y$ ?