Choose the correct answer from the given four options.The solution of the differential equation dy dx + 2xy 1+x^2 =1 (1+x^2)^2 is: 1. y(1+x^2)=C+ ^ -1 x 2. y 1+x^2 =C+ ^ -1 x 3. y (1+x^2)=C+ ^ -1 x 4. y(1+x^2)=C+ ^ -1 x

1. y(1+x^2)=C+ ^ -1 x Solution: Given is, dy dx + 2xy 1+x^2 =1 (1+x^2)^2 Here, P= 2x 1+x^2 and Q=1 (1+x^2)^2 This ia a linerar differential equation. .F.=e^ 2x 1+x^2 dx Put 1+x^2=t 2x dx=dt .F.=e^ dt t =e^ =e^ (1+x^2) =1+x^2 Thus, the general solution is y.(1+x^2)= (1+x^2)1 (1+x^2) +C (1+x^2)= 1 (1+x^2) dx+C (1+x^2)= ^ -1 x+C

Choose the correct answer from the given four options.The solutio…

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Choose the correct answer from the given four options.The solution of the differential equation $\frac{\text{dy}}{\text{dx}}+\frac{2\text{xy}}{1+\text{x}^2}=\frac{1}{(1+\text{x}^2)^2}$ is:

$\text{y}(1+\text{x}^2)=\text{C}+\tan^{-1}\text{x}$
$\frac{\text{y}}{1+\text{x}^2}=\text{C}+\tan^{-1}\text{x}$
$\text{y}\log(1+\text{x}^2)=\text{C}+\tan^{-1}\text{x}$
$\text{y}(1+\text{x}^2)=\text{C}+\sin^{-1}\text{x}$

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Answer

$\text{y}(1+\text{x}^2)=\text{C}+\tan^{-1}\text{x}$

Solution:
Given is, $\frac{\text{dy}}{\text{dx}}+\frac{2\text{xy}}{1+\text{x}^2}=\frac{1}{(1+\text{x}^2)^2}$
Here, $\text{P}=\frac{2\text{x}}{1+\text{x}^2}$ and $\text{Q}=\frac{1}{(1+\text{x}^2)^2}$
This ia a linerar differential equation.
$\therefore\text{I.F.}=\text{e}^{\int\frac{2\text{x}}{1+\text{x}^2}\text{dx}}$
Put $1+\text{x}^2=\text{t}\Rightarrow2\text{x}\text{ dx}=\text{dt}$
$\therefore\text{I.F.}=\text{e}^{\int\frac{\text{dt}}{\text{t}}}=\text{e}^{\log\text{t}}$
$=\text{e}^{\log(1+\text{x}^2)}=1+\text{x}^2$
Thus, the general solution is
$\text{y}.(1+\text{x}^2)=\int(1+\text{x}^2)\frac{1}{(1+\text{x}^2)}+\text{C}$
$\Rightarrow\text{y}(1+\text{x}^2)=\int\frac{1}{(1+\text{x}^2)}\text{dx}+\text{C}$
$\Rightarrow\text{y}(1+\text{x}^2)=\tan^{-1}\text{x}+\text{C}$

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