MCQ
Solution of the differential equation $\sin \frac{{dy}}{{dx}} = a$ with $y(0) = 1$ is
- A${\sin ^{ - 1}}\frac{{(y - 1)}}{x} = a$
- ✓$\sin \frac{{(y - 1)}}{x} = a$
- C$\sin \frac{{(1 - y)}}{{(1 + x)}} = a$
- D$\sin \frac{y}{{(x + 1)}} = a$
Integrating both sides,$\int_{}^{} {dy} = \int_{}^{} {{{\sin }^{ - 1}}a\,dx} $
$y = x{\sin ^{ - 1}}a + c$ and $y(0) = 0 + c = 1$, $\therefore c = 1$
$\therefore y = x{\sin ^{ - 1}}a + 1$ ==> $a = \sin \frac{{y - 1}}{x}$.
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