Question

(4x+3y^2)dx+2xy*dy=0 by using integrating fector

Answer 1

$\underbrace{(4x+3y^2)}_M\,\mathrm dx+\underbrace{2xy}_N\,\mathrm dy=0$

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.