Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDifferential Equations2 Marks
Question
Find the general solution of the differential equation $\frac{d y}{d x}=\sin ^{-1} x$
✓
Answer
Given$\: \frac{d y}{d x}=\sin ^{-1} x$
Separating variables,
$\Rightarrow dy = \sin^{-1} x\ dx$
Integrating both sides,
$\Rightarrow \int d y=\int \sin ^{-1} x\ d x$
Now to integrate $\sin^{-1} x$ we have to multiply it by $1$
because,
$\left.\because(\left\{\int \mathrm{u} . \mathrm{v} \mathrm{dx}=\mathrm{u} \int \mathrm{v} \mathrm{d} \mathrm{x}-\int\left(\frac{\mathrm{d}}{\mathrm{dx}} \mathrm{u}\right)\left(\int \mathrm{vdx}\right) \mathrm{dx}\right\}\mathrm{}\right)$
So,
$\Rightarrow$ y = $\int 1 . \sin ^{-1} x\ d x$
Let $u$ be $\sin^{-1} x$ and $v$ be $1$
We can take the values of $u$ and $v$ from the formula $(I.L.A.T.E)$
$\therefore \mathrm{y}=\left\{\sin ^{-1} \mathrm{x} \int 1 \cdot \mathrm{dx}-\int\left(\frac{\mathrm{d}}{\mathrm{dx}} \sin ^{-1} \mathrm{x}\right)\left(\int 1 \cdot \mathrm{dx}\right) \mathrm{d} \mathrm{x}\right\}$
$\Rightarrow \mathrm{y}=\mathrm{x} \sin ^{-1} \mathrm{x}-\int \frac{\mathrm{x}}{\sqrt{1-\mathrm{x}^{2}}} \mathrm{d} \mathrm{x}$
$\Rightarrow$ let $1 - x^2 = t$
$\Rightarrow - 2x\ dx = dt$
$\Rightarrow x\ dx = -\frac{\mathrm{dt}}{2}$
$\Rightarrow y = x \sin^{-1} x + \int \frac{1}{2 \sqrt{t}} d t$
$\Rightarrow y = x \sin^{-1} x + \frac{1}{2} \int t^{-\frac{1}{2}} d t$
$\Rightarrow y = x \sin^{-1 }x + \frac{1}{2} \sqrt{t}+c$
putting the value of $t,$
$\Rightarrow y = x \sin^{-1} x + \sqrt{1-x^{2}}+c$
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