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Mathematics · 2023 · 3 marks
Differential EquationsMethods of Solving First Order, First Degree Differential Equations3 marksApplyshort answer
CBSE 2023 · Region 1 · Set 3 · Q29
(a) Find the general solution of the differential equation: (d y)/(d x)-(2 y)/x= sin 1/x.
(a)
Find the general solution of the differential equation : \[\frac{d \mathrm{y}}{d \mathrm{x}}-\frac{2 \mathrm{y}}{\mathrm{x}}=\sin \frac{1}{\mathrm{x}} . \]
OR
(b)
Find the particular solution of the differential equation : \[\frac{\mathrm{dy}}{\mathrm{dx}}=\sin (\mathrm{x}+\mathrm{y})+\sin (\mathrm{x}-\mathrm{y}) \text {, given that when } \mathrm{x}=\frac{\pi}{4}, \mathrm{y}=0 \text {. } \]
Marking-scheme solution
Integrating factor $\displaystyle =\mathrm{e}^{\int-\frac{2}{\mathrm{x}} \mathrm{dx}}=\mathrm{e}^{-2 \log \mathrm{x}}=\frac{1}{\mathrm{x}^{2}}$
Solution is $\displaystyle \mathrm{y} \cdot \frac{1}{\mathrm{x}^{2}}=\int \frac{1}{\mathrm{x}^{2}} \cdot \sin \frac{1}{\mathrm{x}} \mathrm{dx}+\mathrm{C}$
Let $\displaystyle \frac{1}{\mathrm{x}}=\mathrm{t},-\frac{1}{\mathrm{x}^{2}} d \mathrm{x}=d \mathrm{t}$
$\displaystyle \therefore \int \frac{1}{\mathrm{x}^{2}} \cdot \sin \frac{1}{\mathrm{x}} \mathrm{dx}=-\int \sin \mathrm{t} \mathrm{dt}=\cos \mathrm{t}=\cos \frac{1}{\mathrm{x}}$
∴ $\displaystyle \mathrm{y} \cdot \frac{1}{\mathrm{x}^{2}}=\cos \frac{1}{\mathrm{x}}+\mathrm{c}$
Given differential equation becomes
$$\frac{d \mathrm{y}}{d \mathrm{x}}=$\displaystyle 2$ \sin \mathrm{x} \cos \mathrm{y}
$$$\Rightarrow \sec \mathrm{y} d \mathrm{y}=2 \sin \mathrm{x} d \mathrm{x}$
Integrating we get
$$\begin{aligned}
& \log |\sec \mathrm{y}+\tan \mathrm{y}|=-$\displaystyle 2$ \cos \mathrm{x}+\mathrm{C}
& \mathrm{x}=\frac{\pi}{4}, \mathrm{y}=$\displaystyle 0$ \text { gives } \mathrm{C}=\sqrt{2}
\end{aligned}
$$∴ particular solution is $\displaystyle \log |\sec \mathrm{y}+\tan \mathrm{y}|=-2 \cos \mathrm{x}+\sqrt{2}$
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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.