CBSE 2022 · Region 3 · Set 1 · Q8 · 3 marks
Find the particular solution of the differential equation $\displaystyle x \frac{\mathrm{dy}}{\mathrm{d} x}-\mathrm{y}=x^{2} \cdot \mathrm{e}^{x}$, given $\displaystyle \mathrm{y}(1)=0$.
OR Find the general solution of the differential equation $\displaystyle x \frac{\mathrm{dy}}{\mathrm{d} x}=\mathrm{y}(\log \mathrm{y}-\log x+1)$.
Marking-scheme solution
\[\frac{dy}{dx} - \frac{y}{x} = \left(\frac{x^{2}}{x}\right)e^{x} = xe^{x}\]
\[I\cdot F = e^{\int -\frac{1}{x}dx} = e^{-\log x} = e^{\log x^{-1}} = \frac{1}{x}\]
Solution is $\displaystyle \displaystyle y \times \frac{1}{x} = \int e^{x}\,dx$
\[\frac{y}{x} = e^{x} + C\]
\[y = xe^{x} + Cx\]
\[C = -e\]Particular solution is
\[y = xe^{x} - ex\]
\[x\frac{dy}{dx} = y[\log y - \log x + 1]\]
\[\frac{dy}{dx} = \frac{y}{x}\left[\log \frac{y}{x} + 1\right]\]
\[y = vx\]
\[\frac{dy}{dx} = v + x\frac{dv}{dx}\]
\[v + x\frac{dv}{dx} = v[\log v + 1]\]
\[x\frac{dv}{dx} = v\log v\]
\[\int \frac{dv}{v\log v} = \int \frac{1}{x}\,dx\]
\[\log|\log v| = \log|x| + \log C\]
\[\log\left(\frac{y}{x}\right) = Cx\]
Differential EquationsMethods of Solving First Order, First Degree Differential EquationsApplyshort_answermedium
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CBSE Class 12 Mathematics past-paper question from the 2022board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.