CBSE 2024 · Region 5 · Set 1 · Q29 · 3 marks
Find the particular solution of the differential equation \[\frac{\mathrm{dy}}{\mathrm{dx}}-2 \mathrm{x} \mathrm{y}=3 \mathrm{x}^{2} \mathrm{e}^{\mathrm{x}^{2}} ; \mathrm{y}(0)=5 \]Solve the following differential equation : \[\mathrm{x}^{2} d \mathrm{y}+\mathrm{y}(\mathrm{x}+\mathrm{y}) d \mathrm{x}=0 \]
Find the particular solution of the differential equation \[\frac{\mathrm{dy}}{\mathrm{dx}}-2 \mathrm{x} \mathrm{y}=3 \mathrm{x}^{2} \mathrm{e}^{\mathrm{x}^{2}} ; \mathrm{y}(0)=5 \]
Solve the following differential equation : \[\mathrm{x}^{2} d \mathrm{y}+\mathrm{y}(\mathrm{x}+\mathrm{y}) d \mathrm{x}=0 \]
Marking-scheme solution
(a)
Given differential equation is a linear order differential equation with:
$$P=-$\displaystyle 2$ \mathrm{x}, Q=$\displaystyle 3$ \mathrm{x}^{$\displaystyle 2$} \mathrm{e}^{\mathrm{x}^{$\displaystyle 2$}}
$$\text { Integrating Factor }=\mathbf{e}^{\int-$\displaystyle 2$ \mathrm{xdx}}=\mathbf{e}^{-\mathrm{x}^{$\displaystyle 2$}}
$$The general solution is: $\displaystyle \mathrm{y} \cdot \mathrm{e}^{-\mathrm{x}^{2}}=\int \mathrm{e}^{-\mathrm{x}^{2}} \cdot 3 \mathrm{x}^{2} \mathrm{e}^{\mathrm{x}^{2}} d \mathrm{x}+C \Rightarrow \mathrm{y} \cdot \mathrm{e}^{-\mathrm{x}^{2}}=\mathrm{x}^{3}+C$
Putting $\displaystyle \mathrm{x}=0, \mathrm{y}=5$, we get, $\displaystyle C=5$
∴ The Particular solution is: $\displaystyle \mathbf{y} \cdot \mathbf{e}^{-\mathbf{x}^{\mathbf{2}}}=\mathbf{x}^{\mathbf{3}}+5$ or $\displaystyle \mathbf{y}=\left(\mathbf{x}^{\mathbf{3}}+5\right) \mathbf{e}^{\mathbf{x}^{\mathbf{2}}}$
Or
(b) $\displaystyle \mathrm{x}^{2} d \mathrm{y}+\mathrm{y}(\mathrm{x}+\mathrm{y}) d \mathrm{x}=0 \Rightarrow \frac{d \mathrm{y}}{d \mathrm{x}}=-\frac{\mathrm{y}}{\mathrm{x}}-\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^{2}$
$$\text { Putting } \frac{\mathrm{y}}{\mathrm{x}}=v \Rightarrow \mathrm{y}=v \mathrm{x}, \frac{d \mathrm{y}}{d \mathrm{x}}=v+\mathrm{x} \frac{d v}{d \mathrm{x}}
\]
\[
v+\mathrm{x} \frac{d v}{d \mathrm{x}}=-v-v^{$\displaystyle 2$}separating the variable and integrating\int \frac{1}{v^{$\displaystyle 2$}+$\displaystyle 2$ v} d v=-\int \frac{1}{\mathrm{x}} d \mathrm{x}
\Rightarrow \int \frac{1}{(v+$\displaystyle 1$)^{$\displaystyle 2$}-$\displaystyle 1$} d v=-\int \frac{1}{\mathrm{x}} d \mathrm{x}$\displaystyle \Rightarrow \frac{1}{2} \log \left|\frac{v}{v+2}\right|=\log \left|\frac{C}{\mathrm{x}}\right|$
Differential EquationsMethods of Solving First Order, First Degree Differential EquationsApplyshort_answermedium
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CBSE Class 12 Mathematics past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.