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DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS1 Mark
Question
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:

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

  2. $\frac{\text{y}}{1+\text{x}^2}=\text{C}+\tan^{-1}\text{x}$

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

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

✓

Answer

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

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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