Question

Consider the differential equation $y - 4 y^{2} = ( y^{7}+1x)y'$ where y(0)=1 Solve this equation by the following method: First, find a suitable integrating factor to obtain an implicit solution F(x,y) = C. This implicit solution cannot be solved explicitly for y but it can be solved explicitly for x.

Answer 1

You have

$\underbrace{(y-4y^2)}_M\,\mathrm dx-\underbrace{(y^7+x)}_N\,\mathrm dy=0$

with partial derivatives

$M_y=1-8y$
$N_x=-1$

An integrating factor in terms of $y$ is easy to find:

$\mu(y)=\exp\left(\displaystyle\int\frac{N_x-M_y}M\,\mathrm dy\right)$
$\mu(y)=\exp\left(\displaystyle\int\frac{8y-2}{y-4y^2}\,\mathrm dy\right)$
$\implies\mu(y)=\dfrac1{y^2}$

Distributing $\mu(y)$ across the ODE gives

$\underbrace{\dfrac{y-4y^2}{y^2}}_{M^*}\,\mathrm dx-\underbrace{\dfrac{y^7+x}{y^2}}_{N^*}\,\mathrm dy=0$

with partial derivatives

${M^*}_y=-\dfrac1{y^2}$
${N^*}_x=-\dfrac1{y^2}$

so the modified ODE is exact.

Now,

$F_x(x,y)=M^*(x,y)$
$F(x,y)=\displaystyle\int\dfrac{y-4y^2}{y^2}\,\mathrm dx$
$F(x,y)=\left(\dfrac1y-4\right)x+f(y)$

$F_y(x,y)=N^*(x,y)$
$-\dfrac x{y^2}+f'(y)=-y^5-\dfrac x{y^2}$
$f'(y)=-y^5$
$\implies f(y)=-\dfrac16y^6+C$

and so the general solution is

$F(x,y)=\left(\dfrac1y-4\right)x-\dfrac{y^6}6=C$

Solving explicitly for $x$, we have

$x=\dfrac{C+\frac{y^6}6}{\frac1y-4}=\dfrac{Cy+y^7}{6-24y}$