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DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS1 Mark
Question
Fill in the blanks.
The solution of $(1+\text{x})^2\frac{\text{dy}}{\text{dx}}+2\text{xy}-4\text{x}^2=0$ is _________.
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Answer

The solution of $(1+\text{x})^2\frac{\text{dy}}{\text{dx}}+2\text{xy}-4\text{x}^2=0$ is $\text{y}=\frac{4\text{x}^3}{(1+\text{x}^2)}+\text{C}(1+\text{x}^2)^{-1}.$
Solution:
Given differential equation is $(1+\text{x})^2\frac{\text{dy}}{\text{dx}}+2\text{xy}-4\text{x}^2=0$
Dividing both sides by $(1 + x^2),$ we get
$\Rightarrow\frac{\text{dy}}{\text{dx}}+\frac{2\text{x}}{1+\text{x}^2}-\frac{4\text{x}^2}{1+\text{x}^2}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}+\frac{2\text{x}}{1+\text{x}^2}\text{y}=\frac{4\text{x}^2}{1+\text{x}^2}$
$\therefore\text{I.F.}=\text{e}^{\frac{2\text{x}}{1+\text{x}^2}\text{dx}}$
Put $1+\text{x}^2=\text{t}\Rightarrow2\text{xdx}=\text{dt}$
$\therefore\text{I.F.}=\text{e}^{\int\frac{\text{dt}}{\text{t}}}=\text{e}^{\log\text{t}}$
$\text{e}^{\log(1+\text{x}^2)}=1+\text{x}^2$
The general solution is
$\text{y}.(1+\text{x}^2)=\int(1+\text{x}^2)\frac{4\text{x}^2}{(1+\text{x}^2)}\text{dx}+\text{C}$
$\Rightarrow(1+\text{x}^2)\text{y}=\int4\text{x}^2\text{dx}+\text{C}$
$\Rightarrow(1+\text{x}^2)\text{y}=4\frac{\text{x}^3}{3}+\text{C}$
$\Rightarrow\text{y}=\frac{4\text{x}^3}{3(1+\text{x}^2)}+\text{C}(1+\text{x}^2)^{-1}$

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