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Solve the following differential equation : $xdy-(y+2x^2)dx=0 $

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  • If the equation is of the form $\large\frac{dy}{dx}$$ + py= Q$, then the integrating factor is $e^{\large\int pdx}$
Step 1:
$xdy -(y+2x^2)dx=0$
Step 2:
Divide throughout by $x$
Clearly this is a homogeneous linear differential equation of the form $\large\frac{dy}{dx}$$+Py=Q$
The integrating factor I.F is $e^{\int Pdx}=e^{\int -\large\frac{1}{x}dx}$
$e^{-\log x}=\large\frac{1}{x}$
Step 3:
The required solution is $ye^{\int Pdx}=\int Qe^{\int Pdx}.dx+c$
(i.e) $y\big(\large\frac{1}{x}\big)=$$\int 2x\big(\large\frac{1}{x}\big)$$dx+c$
$\Rightarrow \large\frac{y}{x}=$$2x+c$
$y-2x^2+cx=0$ is the required solution.
answered Nov 14, 2013 by sreemathi.v
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