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Solve the differential equation : $(1 + x^2)\frac{dy}{dx} +2xy -4x^2 =0$ , subject to the initial condition $y(0) = 0$

Answers (1)

Given : $(1+x^2)\frac{\mathrm{d} y}{\mathrm{d} x}+2xy-4x^2=0$

$\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}+\frac{2xy}{1+x^2}-\frac{4x ^2}{1+x^2}=0$

$\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}+\frac{2xy}{1+x^2}=\frac{4x ^2}{1+x^2}$

$\frac{\mathrm{d} y}{\mathrm{d} x}+Py=Q$

Here : $P=\frac{2x}{1+x^2}$ , $Q=\frac{4x^2}{1+x^2}$

$I.F =e^{\int\! P \: dx}=e^{\int\! \frac{2x}{1+x^2}}=e^{\ln (1+x^2)}=(1+x^2)$

Solution is :

$y(I.F)=\int\! Q\cdot I.F\: dx +c$

$y(1+x^2)=\int\!\frac{4x^2}{1+x^2}\cdot (1+x^2) dx+c$

$y(1+x^2)=4\int\!x^2 dx+c$

$y(1+x^2)=4\frac{x^3}{3}+c\: \: \: \: \: \: -(1)$

Applying condition in (1),

$x=0,y=0$

$0(1+0)=4\left ( \frac{0}{3} \right )+c\Rightarrow c=0$

Putting c value in (1)

$y(1+x^2)=\frac{4x^3}{3}$

or $y=\frac{4x^3}{3(1+x^2)}$

Posted by

Ravindra Pindel

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