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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 1 · Q28
(a) Find the particular solution of the differential equation dy/dx+ sec^2 x · y = tan x · sec^2 x, given that y (0)=0.
(a)
Find the particular solution of the differential equation $\displaystyle \frac{\mathrm{dy}}{\mathrm{dx}}+\sec ^{2} \mathrm{x} \cdot \mathrm{y}=\tan \mathrm{x} \cdot \sec ^{2} \mathrm{x}$, given that $\displaystyle \mathrm{y}(0)=0$.
OR
(b)
Solve the differential equation given by \[\mathrm{x} d \mathrm{y}-\mathrm{y} d \mathrm{x}-\sqrt{\mathrm{x}^{2}+\mathrm{y}^{2}} d \mathrm{x}=0 \]
Marking-scheme solution
Let (a) Integrating factor $\displaystyle =\mathrm{e}^{\int \sec ^{2} \mathrm{x} \mathrm{dx}}=\mathrm{e}^{\tan \mathrm{x}}$
Solution is $\displaystyle \mathrm{ye}^{\tan \mathrm{x}}=\int \tan \mathrm{x} \sec ^{2} \mathrm{x} \mathrm{e}^{\tan \mathrm{x}} \mathrm{dx}+\mathrm{C}$
Let $\displaystyle \tan \mathrm{x}=\mathrm{t} \sec ^{2} \mathrm{x} \mathrm{dx}=\mathrm{dt}$
$$\therefore \int \mathrm{e}^{\tan \mathrm{x}} \tan \mathrm{x} \sec ^{$\displaystyle 2$} \mathrm{x} d \mathrm{x}=\int \mathrm{e}^{\mathrm{t}} \mathrm{t} d \mathrm{t}=\mathrm{e}^{\mathrm{t}}(\mathrm{t}-$\displaystyle 1$)
$$$\therefore \mathrm{ye}^{\tan \mathrm{x}}=\mathrm{e}^{\tan \mathrm{x}}(\tan \mathrm{x}-1)+\mathrm{C}$
$\displaystyle \mathrm{y}(0)=0$ gives $\displaystyle \mathrm{C}=1$
Particular solution is $\displaystyle \mathrm{ye}^{\tan \mathrm{x}}=\mathrm{e}^{\tan \mathrm{x}}(\tan \mathrm{x}-1)+1$ or $\displaystyle \mathrm{y}=\tan \mathrm{x}-1+\mathrm{e}^{-\tan \mathrm{x}}$
Given differential equation can be written as
$$\begin{equation*}
\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{y}}{\mathrm{x}}+\sqrt{$\displaystyle 1$+\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^{$\displaystyle 2$}} \tag{i}
\end{equation*}
$$Let $\displaystyle \mathrm{y}=\mathrm{vx} \Rightarrow \frac{\mathrm{dv}}{\mathrm{dx}}=\mathrm{v}+\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}$ substituting in (i)
We get $\displaystyle \mathrm{v}+\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=\mathrm{v}+\sqrt{1+\mathrm{v}^{2}}$
$$\Rightarrow \frac{d \mathrm{v}}{\sqrt{$\displaystyle 1$+\mathrm{v}^{$\displaystyle 2$}}}=\frac{d \mathrm{x}}{\mathrm{x}}
$$Integrating both sides, we get
$$\log \left|\sqrt{$\displaystyle 1$+\mathrm{v}^{$\displaystyle 2$}}+\mathrm{v}\right|=\log |\mathrm{x}|+\log \mathrm{C}
$$$$
\mathrm{y}+\sqrt{\mathrm{x}^{$\displaystyle 2$}+\mathrm{y}^{$\displaystyle 2$}}=\mathrm{C} \mathrm{x}^{$\displaystyle 2$}
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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.