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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 the general solution:
$\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}-\text{x}+\text{xy}\cot\text{x}=0\ (\text{x}\neq0)$
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Answer

$\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}-\text{x}+\text{xy}\cot\text{x}=0$ $\Rightarrow\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}(1+\text{x}\cot\text{x})=\text{x}$ $\Rightarrow\text{x}\frac{\text{dy}}{\text{dx}}+\Big(\frac{1}{\text{x}}+\cot\text{x}\Big)\text{y}=1$ The equation is a linear differential equation of the form: $\frac{\text{dy}}{\text{dx}}+\text{py}=\text{Q}\ (\text{where p} )=\frac{1}{\text{x}}+\cot\text{x}\ \text{and}\ \text{Q}=1$ $\text{Now, I.F}=\text{e}^{\int\text{pdx}}=\text{e}^{\int\Big(\frac{1}{\text{x}}+\cot\text{x}\Big)\text{dx}}=\text{e}^{\log\text{x}+\log(\sin\text{x})}=\text{e}^{\log(\text{x}\sin\text{x})}=\text{x}\sin\text{x}.$The general solution of the given differential equation is given by the relation,
$\text{y(I.F)}=\int(\text{Q}\times\text{I.F})\text{dx}+\text{C}$ $\Rightarrow\text{y}(\text{x}\sin\text{x})=\int(1\times\text{x}\sin\text{x})\text{dx}+\text{C}$ $\Rightarrow\text{y}(\text{x}\sin\text{x})=\int(\text{x}\sin\text{x})\text{dx}+\text{C}$ $\Rightarrow\text{y}(\text{x}\sin\text{x})=\text{x}\int\sin\text{x}\ \text{dx}-\int\Big[\frac{\text{d}}{\text{dx}}(\text{x}\cdot\int\sin\text{x}\ \text{dx})\Big]+\text{C}$ $\Rightarrow\text{y}(\text{x}\sin\text{x})=\text{x}(-\cos\text{x})-\int1\cdot(-\cos\text{x})\text{dx}+\text{C}$ $\Rightarrow\text{y}(\text{x}\sin\text{x})=-\text{x}\cos\text{x}+\sin\text{x}+\text{C}$ $\Rightarrow\text{y}=\frac{-\text{x}\cos\text{x}}{\text{x}\sin\text{x}}+\frac{\sin\text{x}}{\text{x}\sin\text{x}}+\frac{\text{C}}{\text{x}\sin\text{x}}$ $\Rightarrow\text{y}=-\cot\cdot\text{x}+\frac{1}{\text{x}}+\frac{\text{C}}{\text{x}\sin\text{x}}$

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