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DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS5 Marks
Question
For each of the differential equation given in find the general solution:
$(1+\text{x}^2)\ \text{dy}+2\text{xy}\ \text{dx}=\cot\text{x}\ \text{dx}\ (\text{x}\neq0)$
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Answer

$(1+\text{x}^2)\ \text{dx}+2\text{xy}\ \text{dx}=\cot\text{xdx}$ $\Rightarrow\ \frac{\text{dy}}{\text{dx}}+\frac{2\text{xy}}{1+\text{x}^2}=\frac{\cot\text{xdx}}{1+\text{x}^2}$This equation is a linear dyfferential equation of the form:
$\frac{\text{dy}}{\text{dx}}+\text{py}=\text{Q}\ \big(\text{where p}=\frac{2\text{x}}{1+\text{x}^2}\ \text{and}\ \text{Q}=\frac{\cot\text{x}}{1+\text{x}^2}\big)$ $\text{Now, I.F}=\text{e}^{\int{\text{pdx}}}=\text{e}^{\int{\frac{2\text{x}}{1+\text{x}^2}\text{dx}}}=\text{e}^{\log(1+\text{x}^2)}=1+\text{x}^2.$ The general solution of the given differential equation is given by the relation, $\text{y}(\text{I.F})=\int(\text{Q}\times\text{I.F})\ \text{dx}+\text{C}$ $\Rightarrow\ ​​\text{y}(1+\text{x}^2)=\int\Big[\frac{\cot\text{x}}{1+\text{x}}\times(1+\text{x}^2)\Big]\text{dx}+\text{C}$ $\Rightarrow\ \text{y}(1+\text{x}^2)=\int\cot\text{x dx}+\text{C}$ $\Rightarrow\ \text{y}(1+\text{x}^2)=\log|\sin\text{x}|+\text{C}$

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