Question

sin^-1(dy / dx) = x + y

Answer 1

$$\sin ^{-1}\left(\frac{d y}{d x}\right)=x+y\\ \therefore \sin (x+y)=\frac{d y}{d x}\\ \text{Let } x+y=u .\\ \therefore \frac{d}{d x}(x+y)=\frac{d}{d x}(u),\\ i . e ., 1+\frac{d y}{d x}=\frac{d u}{d x}.\\ \text{Subst.ing in the diff. eqn.,}\\ \sin u=\frac{d u}{d x}-1, or,\frac{d u}{d x}=1+\sin u .\\ \therefore \frac{d u}{1+\sin u}=d x \ldots \ldots \ldots \ldots . . \text{ [Separable Variable]}.\\ \therefore \int \frac{d u}{1+\sin u}=\int d x+c\\ \therefore \int \frac{1-\sin u}{1-\sin ^{2} u} d u=x+c\\ or, \int \frac{1-\sin u}{\cos ^{2} u} d u=x+c.\\ \therefore \int\left\{\frac{1}{\cos ^{2} u}-\frac{\sin u}{\cos ^{2} u}\right\} d u=x+c .\\ \therefore \tan u-\sec u=x+c\\ \text{ Letting } u=x+y, \text{ we get the general solution as under: }\\ \tan (x+y)-\sec (x+y)=x+c\\$$