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Steve Pugh · Answer 1 · 2018-01-19T04:27:46+0000

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

This equation is not exact, since

M_y=-84x^3

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

$\mu_yM+\mu M_y=\mu_xN+\mu N_x$

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

When we take
$\mu$ to be a function of either

or

, but not both, this partial differential equation reduces to one of the separable ordinary differential equations

$\frac{\mu_y}\mu=\frac{N_x-M_y}M$

$\frac{\mu_x}\mu=-\frac{N_x-M_y}N$

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

or

, 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

$\frac{N_x-M_y}M=(252x^3y+84x^3)/(-84x^3y-28x^3)=-3$

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

, which means
$\mu_x=0$ , and it satisfies

$\frac{\mu_y}\mu=\frac1\mu(\mathrm d\mu)/(\mathrm dy)=-3\implies\displaystyle\int\frac{\mathrm d\mu}\mu=-3\int\mathrm dy$

$\implies\ln\mu=-3y\implies\mu=e^(-3y)$

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