Solve the equation (35 x^(4) y+14 x^(5) y-2 y^(3)-4x y^(3) )dx + (7 x^(5) -6x y^(2) )dy leaving your answer in implicit form.

Question

Solve the equation
`(35 x^(4) y+14 x^(5) y-2 y^(3)-4x y^(3) )dx + (7 x^(5) -6x y^(2) )dy` leaving your answer in implicit form.

Answer 1

`\underbrace{35x^4y+14x^5y-2y^3-4xy^3}_M\,\mathrm dx+\underbrace{7x^5-6xy^2}_N\,\mathrm dy=0`


`M_y=35x^4+14x^5-6y^2-12xy^2`

`N_x=35x^4-6y^2`


`\frac{N_x-M_y}N=(-14x^5+12xy^2)/(7x^5-6xy^2)=-2`

This suggests an integrating factor depending on
`x` only is possible, and given by


`\mu(x)=\exp\left(-\displaystyle\int\frac{N_x-M_y}N\,\mathrm dx\right)=e^(2x)`

Distributing across the ODE, we end up with


`\underbrace{(35x^4y+14x^5y-2y^3-4xy^3)e^(2x)}_(M^*)\,\mathrm dx+\underbrace{(7x^5-6xy^2)e^(2x)}_(N^*)\,\mathrm dy=0`

The equation is now exact, with


`{M^*}_y={N^*}_x=(35x^4+14x^5-6y^2-12xy^2)e^(2x)`

Now we find the solution:


`F_x=M^*`

`F=\displaystyle\int(35x^4+14x^5-2y^3-4xy^3)e^(2x)\,\mathrm dx`

`F=(7x^5y-2xy^3)e^(2x)+f(y)`


`F_y=N^*`

`(7x^5-6xy^2)e^(2x)+f'(y)=(7x^5-6xy^2)e^(2x)`

`f'(y)=0`

`\implies f(y)=C`

The general solution is then


`F(x,y)=(7x^5y-2xy^3)e^(2x)=C`