Solve the differential equation: (i) `(1+y^2) + ( x - e^(tan^-1 y) ) dy/dx = 0` (ii) ` x dy/dx + cos^2 y = tan y dy/dx`

Explanation

Given Differential equation can be written as

`(1+y^{2}) \frac{d x}{d y}+x=e^{\tan ^{-1}} y`

or `\frac{d x}{d y}+\frac{1}{1+y^{2}} . x=\frac{e^{\tan ^{-1}} y}{1+y^{2}}`

Here, `I.F. =e^{\int \frac{1}{1+y^{2} d y}}=e^{\tan ^{-1} y}`

Solution is `x e^{\tan ^{-1} y}=\int \frac{e^{\tan ^{-1} y}}{1+y^{2}} e^{\tan ^{-1} y} d y`

`Rightarrow x e^{\tan ^{-1} y}=\frac{e^{2 \tan ^{-1} y}}{2}+C`