\(\underbrace{(4x+3y^2)}_M\,\mathrm dx+\underbrace{2xy}_N\,\mathrm dy=0\)
The ODE is exact if \(\dfrac{\partial M}{\partial y}=\dfrac{\partial N}{\partial x}\).
\(M_y=6y\)
\(N_x=2y\)
This is not the case, so look for an integrating factor \(\mu(x)\) such that
\(\dfrac\partial{\partial y}\mu M=\dfrac\partial{\partial x}\mu N\)
Since \(\mu\) is a function of \(x\) only, you have
\(\mu M_y=\mu'N+\mu N_x\implies\dfrac{\mu'}\mu=\dfrac{M_y-N_x}N\implies\mu=\exp\left(\displaystyle\int\frac{M_y-N_x}N\,\mathrm dx\right)\)
So, the integrating factor is
\(\mu=\exp\left(\displaystyle\int\frac{6y-2y}{2xy}\,\mathrm dx\right)=\exp\left(2\int\frac{\mathrm dx}x\right)=x^2\)
Now the ODE can be modified as
\(\underbrace{(4x^3+3x^2y^2)}_{M^*}\,\mathrm dx+\underbrace{2x^3y}_{N^*}\,\mathrm dy=0\)
Check for exactness:
\({M^*}_y=6x^2y\)
\({N^*}_x=6x^2y\)
so the modified ODE is indeed exact.
Now, you're looking for a solution of the form \(\Psi(x,y)=C\), since differentiating via the chain rule yields
\(\dfrac{\mathrm d}{\mathrm dx}\Psi(x,y)=\Psi_x+\Psi_y\dfrac{\mathrm dy}{\mathrm dx}=0\)
Matching up components, you would have
\(\Psi_x=M^*=4x^3+3x^2y^2\)
\(\displaystyle\int\Psi_x\,\mathrm dx=\int(4x^3+3x^2y^2)\,\mathrm dx\)
\(\Psi=x^4+x^3y^2+f(y)\)
Differentiate this with respect to \(y\) to get
\(\Psi_y=2x^3y+f'(y)=2x^3y=N^*\)
\(f'(y)=0\implies f(y)=C_1\)
So the solution here is
\(\Psi(x,y)=x^4+x^3y^2+C_1=C\implies x^4+x^3y^2=C\)
Just for a final check, take the derivative to get back the original ODE:
\(\dfrac{\mathrm d}{\mathrm dx}[x^4+x^3y^2]=\dfrac{\mathrm d}{\mathrm dx}C\)
\(4x^3+3x^2y^2+2x^3y\dfrac{\mathrm dy}{\mathrm dx}=0\)
\(4x+3y^2+2xy\dfrac{\mathrm dy}{\mathrm dx}=0\)
\((4x+3y^2)\,\mathrm dx+2xy\,\mathrm dy=0\)
so the solution is correct.
