Question

10. (10 points) Solve the following ODE. (3y^2 + sec^2 (x)) dx +(6xy + y) dy = 0

Answer 1

The ODE is exact, since

$(3y^2+\sec^2x)_y=6y$

$(6xy+y)_x=6y$

so there is a solution of the form $\Psi(x,y)=C$ satisfying

$\Psi_x=3y^2+\sec^2x$

$\Psi_y=6xy+y$

Integrating both sides of the first PDE wrt $x$ gives

$\Psi=3xy^2+\tan x+f(y)$

and differentiating wrt $y$ gives

$\Psi_y=6xy+y=6xy+f'(y)\implies f'(y)=y\implies f(y)=\dfrac{y^2}2+C$

Then

$\Psi(x,y)=\boxed{3xy^2+\tan x+\dfrac{y^2}2=C}$