Gujarat 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)
$\therefore {y}=\left\{\sin ^{-1} x \int 1 \cdot dx-\int\left(\frac{d}{dx} \sin ^{-1} x\right)\left(\int 1 \cdot dx\right) d x\right\}$
$\Rightarrow {y}=x \sin ^{-1} x-\int \frac{x}{\sqrt{1-x^{2}}} d x$
$\Rightarrow$ let $1 - x^2 = t$
$\Rightarrow$ - 2x dx = dt
$\Rightarrow xdx = -\frac{\mathrm{dt}}{2}$
$\Rightarrow y = x \sin^{-1} x + \int \frac{1}{2 \sqrt{t}} d t$
$\Rightarrow y = x \sin^{-1} x + \frac{1}{2} \int t^{-\frac{1}{2}} d t$
$\Rightarrow y = x \sin^{-1}x + \frac{1}{2} \sqrt{t}+c$
putting the value of t,
$\Rightarrow y = x \sin^{-1} x + \sqrt{1-x^{2}}+c$
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