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DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS3 Marks
Question
For each of the differential equation given in find the general solution: $\text{x}\log\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=\frac{2}{\text{x}}\log\text{x}$
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Answer

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

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