The solution of the equation sin^{-1} left( f | vedclass

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Question

(B) Given the differential equation $\sin^{-1} \left( \frac{dy}{dx} \right) = x + y$. Taking sine on both sides,we get $\frac{dy}{dx} = \sin(x + y)$.

Vedclass · Accepted Answer

$	an(x + y) - \sec(x + y) = x + c$

Vedclass · Answer

(B) Given the differential equation $\sin^{-1} \left( \frac{dy}{dx} ight) = x + y$. Taking sine on both sides,we get $\frac{dy}{dx} = \sin(x + y)$. Let $v = x + y$. Then,differentiating with respect to $x$,we get $\frac{dv}{dx} = 1 + \frac{dy}{dx}$,which implies $\frac{dy}{dx} = \frac{dv}{dx} - 1$. Substituting these into the equation,we have $\frac{dv}{dx} - 1 = \sin v$,or $\frac{dv}{dx} = 1 + \sin v$. Separating the variables,we get $\frac{dv}{1 + \sin v} = dx$. Multiplying the numerator and denominator by $(1 - \sin v)$,we get $\frac{1 - \sin v}{1 - \sin^2 v} dv = dx$,which simplifies to $\frac{1 - \sin v}{\cos^2 v} dv = dx$. This can be written as $(\sec^2 v - \sec v 	an v) dv = dx$. Integrating both sides,we get $\int (\sec^2 v - \sec v 	an v) dv = \int dx$. This results in $	an v - \sec v = x + c$. Substituting $v = x + y$ back,we get $	an(x + y) - \sec(x + y) = x + c$.

The solution of the equation $\sin^{-1} \left( \frac{dy}{dx} \right) = x + y$ is

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