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Mathematics · 2023 · 3 marks
Differential EquationsMethods of Solving First Order, First Degree Differential Equations3 marksApplyshort answer
CBSE 2023 · Region 4 · Set 1 · Q29
(a) Find the particular solution of the differential equation dy/dx=(x + y)/x, y (1)=0.
(a)
Find the particular solution of the differential equation \[\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{x}+\mathrm{y}}{\mathrm{x}}, \mathrm{y}(1)=0 . \]
OR
(b)
Find the general solution of the differential equation \[\mathrm{e}^{\mathrm{x}} \tan \mathrm{y} \mathrm{dx}+\left(1-\mathrm{e}^{\mathrm{x}}\right) \sec ^{2} \mathrm{y} \mathrm{dy}=0 . \]
Marking-scheme solution
$\displaystyle \frac{d \mathrm{y}}{d \mathrm{x}}=\frac{\mathrm{x}+\mathrm{y}}{\mathrm{x}} \Rightarrow \frac{d \mathrm{y}}{d \mathrm{x}}=1+\frac{\mathrm{y}}{\mathrm{x}}$
$$\text { Let } \frac{\mathrm{y}}{\mathrm{x}}=\mathrm{v} . \text { Then } \mathrm{x} \frac{d \mathrm{v}}{d \mathrm{x}}+\mathrm{v}=\frac{d \mathrm{y}}{d \mathrm{x}}
$$So, Differential equation becomes.
$$\mathrm{x} \frac{d \mathrm{v}}{d \mathrm{x}}+\mathrm{v}=$\displaystyle 1$+\mathrm{v}
$$$\Rightarrow \mathrm{dv}=\frac{d \mathrm{x}}{\mathrm{x}}$
$$\Rightarrow \mathrm{v}=\log |\mathrm{x}|+\mathrm{c}
$$$$
\Rightarrow \mathrm{y}=\mathrm{x} \log |\mathrm{x}|+\mathrm{cx}
$$$$
\Rightarrow \mathrm{x}=$\displaystyle 1$, \mathrm{y}=$\displaystyle 0$ \Rightarrow \mathrm{c}=$\displaystyle 0$, \mathrm{y}=\mathrm{x} \log |\mathrm{x}|
$$[can also be solved, taking first order linear diff. eqn]
The given D.E. is
$$\frac{\sec ^{$\displaystyle 2$} \mathrm{y}}{\tan \mathrm{y}} d \mathrm{y}=-\frac{\mathrm{e}^{\mathrm{x}}}{$\displaystyle 1$-\mathrm{e}^{\mathrm{x}}} d \mathrm{x}
$$Integrating
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CBSE Class 12 Mathematics past-paper question from the 2023 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.