Question

Solve (1-sinxtany)dx + (cosxsec^2y)dy = 0

Answer 1

This ODE is exact, since


`\frac\partial{\partial y}[1-\sin x\tan y]=-\sin x\sec^2y`

`\frac\partial{\partial x}[\cos x\sec^2y]=-\sin x\sec^2y`

We're looking for a solution of the form
`\Psi(x,y)=C` such that, by the chain rule,


`(\mathrm d)/(\mathrm dx)\Psi(x,y)=(\partial\Psi)/(\partial x)\,\mathrm dx+(\partial\Psi)/(\partial y)\,\mathrm dy=0`

`\implies\begin{cases}\Psi_x=1-\sin x\tan y\\\Psi_y=\cos x\sec^2y\end{cases}`

Integrating the first equation with respect to
`x`, we have


`\displaystyle\int\Psi_x\,\mathrm dx=\int(1-\sin x\tan y)\,\mathrm dx`

`\Psi(x,y)=x+\cos x\tan y+f(y)`

Differentiating with respect to
`y` yields


`\Psi_y=\cos x\sec^2y+f'(y)`

`\cos x\sec^2y=\cos x\sec^2y+f'(y)`

`f'(y)=0`

`\implies f(y)=C`

So the solution to this ODE is


`\Psi(x,y)=x+\cos x\tan y+C=C`

or simply


`\Psi(x,y)=x+\cos x\tan y=C`