The solution of the differential equation sin | vedclass

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(B) Given the differential equation: $\sin ^{-1}\left(\frac{dy}{dx}\right)=x+y$ $\implies \frac{dy}{dx}=\sin (x+y)$ Let $x+y=t$. Then differentiating

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

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(B) Given the differential equation: $\sin ^{-1}\left(\frac{dy}{dx}ight)=x+y$ $\implies \frac{dy}{dx}=\sin (x+y)$ Let $x+y=t$. Then differentiating with respect to $x$,we get $1+\frac{dy}{dx}=\frac{dt}{dx}$,so $\frac{dy}{dx}=\frac{dt}{dx}-1$. Substituting this into the equation: $\frac{dt}{dx}-1=\sin t$ $\implies \frac{dt}{dx}=1+\sin t$ $\implies \int \frac{dt}{1+\sin t}=\int dx$ To integrate $\frac{1}{1+\sin t}$,multiply numerator and denominator by $(1-\sin t)$: $\int \frac{1-\sin t}{1-\sin^2 t} dt = \int dx$ $\implies \int \frac{1-\sin t}{\cos^2 t} dt = x+c$ $\implies \int (\sec^2 t - \sec t 	an t) dt = x+c$ $\implies 	an t - \sec t = x+c$ Substituting $t=x+y$ back,we get: $x = 	an (x+y) - \sec (x+y) + c$.

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

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