Find the equation of a curve passing through origin and satisfying the differential equation $(1 + x^{2}) \frac{d y}{d x} + 2 x y = 4 x^{2}$

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Solution

Given that, $(1 + x^{2}) \frac{d y}{d x} + 2 x y = 4 x^{2}$
$\Rightarrow \frac{d y}{d x} + \frac{2 x}{1 + x^{2}} . y = \frac{4 x^{2}}{1 + x^{2}}$
which is a linear differential equation.
On comparing it with $\frac{d y}{d x} + P y = Q,$ we get
$P = \frac{2 x}{1 + x^{2}}, Q = \frac{4 x^{2}}{1 + x^{2}} ∴ I F = e^{\int P d x} = e^{\int \frac{2 x}{1 + x^{2}}} d x$
$P u t 1 + x^{2} = t \Rightarrow 2 x d x = d t I F = 1 + x^{2} = e^{\int \frac{d t}{t}} = e^{l o g t} = e^{l o g (1 + x^{2})}$
The general solution is
$y . (1 + x^{2}) = \int \frac{4 x^{2}}{1 + x^{2}} (1 + x^{2}) d x + C \Rightarrow y . (1 + x^{2}) = \int 4 x^{2} d x + C \Rightarrow y . (1 + x^{2}) = 4 \frac{x^{3}}{3} + C$ ...(i)
Since, the curve passes through origin, then substituting
x=0 and y=0 in Eq.(i), we get C = 0

So, the required equation of curve is $y (1 + x^{2}) = \frac{4 x^{3}}{3} \Rightarrow y = \frac{4 x^{3}}{3 (1 + x^{2})}$

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Similar questions

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

The equation of the curve passing through the origin and satisfying the equation $(1 + x^{2}) \frac{d y}{d x} + 2 x y = 4 x^{2}$ is

(1 + x^{2}) \frac{d y}{d x} + 2 x y = \frac{1}{(1 + x^{2})}

\frac{d y}{d x} = (x - y)^{2}

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