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(8y-2)/(y-4y^2)\,\mathrm dy\right)`

`\implies\mu(y)=\frac1{y^2}`

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


`\underbrace{(y-4y^2)/(y^2)}_(M^*)\,\mathrm dx-\underbrace{(y^7+x)/(y^2)}_(N^*)\,\mathrm dy=0`

with partial derivatives


`{M^*}_y=-\frac1{y^2}`

`{N^*}_x=-\frac1{y^2}`

so the modified ODE is exact.

Now,


`F_x(x,y)=M^*(x,y)`

`F(x,y)=\displaystyle\int(y-4y^2)/(y^2)\,\mathrm dx`

`F(x,y)=\left(\frac1y-4\right)x+f(y)`


`F_y(x,y)=N^*(x,y)`

`-\frac x{y^2}+f'(y)=-y^5-\frac x{y^2}`

`f'(y)=-y^5`

`\implies f(y)=-\frac16y^6+C`

and so the general solution is


`F(x,y)=\left(\frac1y-4\right)x-\frac{y^6}6=C`

Solving explicitly for
`x`, we have


`x=\frac{C+\frac{y^6}6}{\frac1y-4}=(Cy+y^7)/(6-24y)`