Question

Solve the implicit ODE 7B4%7D%20y-6%29dy%3D0" id="TexFormula1" title="(-84 x^{3} y - 28 x^{3} )dx+(63 x^{4} y-6)dy=0" alt="(-84 x^{3} y - 28 x^{3} )dx+(63 x^{4} y-6)dy=0" align="absmiddle" class="latex-formula"> by finding an integrating factor m that is a either a function m(x) of x or a function m(y) of y only. Now multiply the equation by the integrating factor that you have found and then integrate the resulting equation to get a solution in implicit form.

Answer 1

$\underbrace{(-84x^3y-28x^3)}_{M(x,y)}\,\mathrm dx+\underbrace{(63x^4y-6)}_{N(x,y)}=0$

This equation is not exact, since

$M_y=-84x^3$
$N_x=252x^3y$

So we look for an integrating factor $\mu(x,y)$ such that

$\mu M\,\mathrm dx+\mu N\,\mathrm dy=M^*\,\mathrm dx+N^*\,\mathrm dy=0$

For this to be exact, we require

${M^*}_y={N^*}_x$

Differentiating both sides gives

$\mu_yM+\mu M_y=\mu_xN+\mu N_x$
$\dfrac{\mu_y}\mu M-\dfrac{\mu_x}\mu N=N_x-M_y$

When we take $\mu$ to be a function of either $x$ or $y$, but not both, this partial differential equation reduces to one of the separable ordinary differential equations

$\dfrac{\mu_y}\mu=\dfrac{N_x-M_y}M$
$\dfrac{\mu_x}\mu=-\dfrac{N_x-M_y}N$

both of which can be solved directly for $\mu$ provided that the result on the right hand side of either ODE is a function of either only $y$ or $x$, respectively.

The choice of which integrating factor $\mu$ to look for is then decided by how easily the right hand side can be taken care of. We have

$\dfrac{N_x-M_y}M=\dfrac{252x^3y+84x^3}{-84x^3y-28x^3}=-3$

On the other hand, the integral resulting from an integrating factor $\mu(x)$ is more complicated/impossible to deal with. So, the integrating factor must be a function of $y$, which means $\mu_x=0$, and it satisfies

$\dfrac{\mu_y}\mu=\dfrac1\mu\dfrac{\mathrm d\mu}{\mathrm dy}=-3\implies\displaystyle\int\frac{\mathrm d\mu}\mu=-3\int\mathrm dy$
$\implies\ln\mu=-3y\implies\mu=e^{-3y}$

Distributing the integrating factor across the original ODE, we have

$\underbrace{(-84x^3y-28x^3)e^{-3y}}_{M^*(x,y)}\,\mathrm dx+\underbrace{(63x^4y-6)e^{-3y}}_{N^*(x,y)}=0$

with partial derivatives

${M^*}_y=252x^3ye^{-3y}$
${N^*}_x=252x^3ye^{-3y}$

Thus the modified ODE is exact, as required. Now we try to find a solution of the form $F(x,y)=C$.

$F_x=M^*$
$F=\displaystyle\int(-84x^3y-28x^3)e^{-3y}\,\mathrm dx$
$F=-7x^4(3y+1)e^{-3y}+f(y)$

$F_y=N^*$
$63x^4ye^{-3y}+f'(y)=(63x^4y-6)e^{-3y}$
$f'(y)=-6e^{-3y}$
$f(y)=2e^{-3y}+C$

Therefore the general solution is

$F(x,y)=-7x^4(3y+1)e^{-3y}+2e^{-3y}=C$