1 Answer

Onick · Answer 1 · 2020-01-26T20:06:30+0000

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
gives

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

and differentiating wrt
gives

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

Then

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

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