Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDifferential Equations5 Marks
Question
Solve the following differential equation:$\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=\text{x}\text{e}^{\text{x}}$
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Answer
Here, $\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=\text{x}\text{e}^{\text{x}}$ $\frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}}=\text{e}^{\text{x}}$ It is a linear differential equation, comparing it with $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$ $\text{P}=\frac{1}{\text{x}}, \text{Q}=\text{e}^{\text{x}}$ I.F. $=\text{e}^{\int\text{Pdx}}$ $=\text{e}^{\int\frac{1}{\text{x}}\text{dx}}$ $=\text{e}^{\log\text{x}}$ $=\text{x}$Solution of the equation is given by,
$\text{y}\times(\text{I.F.})=\int\text{Q}\times(\text{I.F.})\text{dx + C}$ $\text{y}\times(\text{x})=\int\text{e}^{\text{x}}\times\text{xdx + C}$ $\text{xy}=\text{x}\int\text{e}^{\text{x}}\text{dx}-\int(1\times\int\text{e}^{\text{x}}\text{dx})\text{dx + C}$ Using integration by parts $=\text{x}\text{e}^{\text{x}}-\int\text{e}^{\text{x}}\text{dx}+\text{C}$ $=\text{x}\text{e}^{\text{x}}-\text{e}^{\text{x}}+\text{C}$ $\text{xy}=(\text{x}-1)\text{e}^{\text{x}}+\text{C}$ $\text{y}=\Big(\frac{\text{x}-1}{\text{x}}\Big)\text{e}^{\text{x}}+\frac{\text{C}}{\text{x}},\text{x}>0$
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