Question

Determine whether the given differential equation is exact. if it is exact, solve it. (if it is not exact, enter not.) (3x + 6y) dx + (6x − 8y3) dy = 0

Answer 1

$\underbrace{(3x+6y)}_{M(x,y)}\,\mathrm dx+\underbrace{(6x-8y^3)}_{N(x,y)}\,\mathrm dy=0$

The ODE is exact if $\dfrac{\partial M}{\partial y}=\dfrac{\partial N}{\partial x}$. We have

$\dfrac{\partial M}{\partial y}=6$
$\dfrac{\partial N}{\partial x}=6$

so the equation is indeed exact.

We're looking for a solution of the form

$\Psi(x,y)=C$

By the chain rule, we have

$\dfrac{\partial\Psi}{\partial x}+\dfrac{\partial\Psi}{\partial y}\dfrac{\mathrm dy}{\mathrm dx}=0$
$\iff\dfrac{\partial\Psi}{\partial x}\,\mathrm dx+\dfrac{\partial\Psi}{\partial y}\,\mathrm dy=0$

which means $\dfrac{\partial\Psi}{\partial x}=M$ and $\dfrac{\partial\Psi}{\partial y}=N$.

Now

$\dfrac{\partial\Psi}{\partial x}=M\implies\displaystyle\int\frac{\partial\Psi}{\partial x}\,\mathrm dx=\int M\,\mathrm dx$
$\implies\Psi(x,y)=\dfrac32x^2+6xy+f(y)$

Differentiating with respect to $y$ yields

$\dfrac{\partial\Psi}{\partial y}=N\implies 6x+f'(y)=6x-8y^3$
$\implies f'(y)=-8y^3\implies f(y)=\displaystyle\int(-8y^3)\,\mathrm dy=-2y^4+C$

and so the general solution is

$\Psi(x,y)=\dfrac32x^2+6xy-2y^4+C=C$
$\implies\Psi(x,y)=\dfrac32x^2+6xy-2y^4=C$