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DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS5 Marks
Question
For each of the differential equations given in find a particular solution satisfying the given condition:
$(1+\text{x}^2)\frac{\text{dy}}{\text{dx}}+2\text{xy}=\frac{1}{1+\text{x}^2};\text{y}=0\ \text{when x}=1$
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Answer

$(1+\text{x}^2)\frac{\text{dy}}{\text{dx}}+2\text{xy}=\frac{1}{1+\text{x}^2}$ $\Rightarrow\frac{​​\text{dy}}{\text{dx}}+\frac{2\text{xy}}{1+\text{x}^2}=\frac{1}{(1+\text{x}^2)^2}$ This is a linear differential equation of the form: $\frac{\text{dy}}{\text{dx}}+\text{py}=\text{Q}\ \Big(\text{where p}=\frac{2\text{x}}{1+\text{x}^2}\ \text{and}\ \text{Q}=\frac{1}{(1+\text{x}^2)^2}\Big)$ $\text{Now, I.F}=\text{e}^{\int\text{pdy}}=\text{e}^{\int\frac{2\text{x dx}}{1+\text{x}^2}}=\text{e}^{\log(1+\text{x}^2)}=1+\text{x}^2.$ The general solution of the given differential equation is given by the relation, $\text{y(I.F)}=\int(\text{Q}\times\text{I.F.})\text{dx}+\text{C}$ $\Rightarrow\text{y}(1+​\text{x}^2​)=\int\bigg[\frac{1}{(1+\text{x}^2)^2}\cdot(1+\text{x}^2)\bigg]\text{dx}+\text{C}$ $\Rightarrow\text{y}(1+\text{x}^2)=\int\frac{1}{1+\text{x}^2}\text{dx}+\text{C}$ $\Rightarrow\text{y}(1+\text{x}^2)=\tan^{-1}\text{x}+\text{C}\ \ ...(1)$ Now, y = 0 at x =1. Therefore, $0=\tan^{-1}1+\text{C}$$\Rightarrow\text{C}=-\frac{\pi}{4}$
$\text{Substituting C}=-\frac{\pi}{4}\ \text{in equation (1), we get:}$$\text{y}(1+\text{x}^2)=\tan^{-1}\text{x}-\frac{\pi}{4}$
This is the required general solution of the given differential equation.

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