The solution of differential equation d y-d x-2 x y-1-x2-1-1-x22 isa y 1- x 2- C -tan -1 xb y-1-x2-C-tan -1 xc ylog1-x2-C-tan -1 xd y1-x2-C-sin -1 x

Given that, dy/dx+2xy/1+x^2=1/(1+x^2)^2 Here, P=2x/1+x^2 and Q=1/(1+x^2)^2 which is a linear differential equation. ∴ IF=e^∫2x/1+x^2dx Put 1+x^2=t ⇒ 2x dx=dt ...

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Byju's Answer

Standard XII

Mathematics

Homogeneous Differential Equation

The solution ...

Question

The solution of differential equation $\frac{d y}{d x} + \frac{2 x y}{1 + x^{2}} = \frac{1}{(1 + x^{2})^{2}}$ is
(a) $y (1 + x^{2}) = C + t a n^{- 1} x$
(b) $\frac{y}{1 + x^{2}} = C + t a n^{- 1} x$
(c) $y l o g (1 + x^{2}) = C + t a n^{- 1} x$
(d) $y (1 + x^{2}) = C + s i n^{- 1} x$

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Solution

Given that, $\frac{d y}{d x} + \frac{2 x y}{1 + x^{2}} = \frac{1}{(1 + x^{2})^{2}}$
Here, $P = \frac{2 x}{1 + x^{2}}$ and $Q = \frac{1}{(1 + x^{2})^{2}}$
which is a linear differential equation.
$∴ I F = e^{\int \frac{2 x}{1 + x^{2}} d x}$
Put $1 + x^{2} = t \Rightarrow 2 x d x = d t ∴ I F = e^{\int \frac{d t}{t}} = e^{l o g t} = e^{l o g (1 + x^{2})} 1 + x^{2}$
The general solution is
$y . (1 + x^{2}) = \int (1 + x^{2}) \frac{1}{(1 + x^{2})^{2}} + C \Rightarrow y (1 + x^{2}) = \int \frac{1}{1 + x^{2}} d x + C \Rightarrow y (1 + x^{2}) = t a n^{- 1} x + C$

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The solution of differential equation dydx+2xy1+x2=1(1+x2)2 is (a) y(1+x2)=C+tan−1x (b) y1+x2=C+tan−1x (c) ylog(1+x2)=C+tan−1x (d) y(1+x2)=C+sin−1x

Given that, dydx+2xy1+x2=1(1+x2)2 Here, P=2x1+x2 and Q=1(1+x2)2 which is a linear differential equation. ∴IF=e∫2x1+x2dx Put 1+x2=t⇒2xdx=dt∴IF=e∫dtt=elogt=elog(1+x2)1+x2 The general solution is y.(1+x2)=∫(1+x2)1(1+x2)2+C⇒y(1+x2)=∫11+x2dx+C⇒y(1+x2)=tan−1x+C

The solution of differential equation $\frac{d y}{d x} + \frac{2 x y}{1 + x^{2}} = \frac{1}{(1 + x^{2})^{2}}$ is
(a) $y (1 + x^{2}) = C + t a n^{- 1} x$
(b) $\frac{y}{1 + x^{2}} = C + t a n^{- 1} x$
(c) $y l o g (1 + x^{2}) = C + t a n^{- 1} x$
(d) $y (1 + x^{2}) = C + s i n^{- 1} x$