Question

Consider the implicit differential equation
`(49 y^(3) + 45 xy) dx + (98 xy^(2) +50 x^(2) )dy=0`. For the integrating factor
`x^(p) y^(q)` of this equation, find p and q. Now multiply the equation by the integrating factor
`x^(p) y^(q)` that you have found and then integrate the resulting equation to get a solution in implicit form.

Answer 1

We're looking for an integrating factor
`\mu(x,y)=x^py^q` such that


`\mu\underbrace{(49y^3+45xy)}_M\,\mathrm dx+\mu\underbrace{(98xy^2+50x^2)}_N\,\mathrm dy=0`

is exact, which would require that


`(\mu M)_y=(\mu N)_x`

`(49x^py^(q+3)+45x^(p+1)y^(q+1))_y=(98x^(p+1)y^(q+2)+50x^(p+2)y^q)_x`

`49(q+3)x^py^(q+2)+45(q+1)x^(p+1)y^q=98(p+1)x^py^(q+2)+50(p+2)x^(p+1)y^q`

`\implies\begin{cases}49(q+3)=98(p+1)\\45(q+1)=50(p+2)\end{cases}\implies p=\frac52,q=4`

You can verify that
`(\mu M)_y=(\mu N)_x` if you'd like. With the ODE now exact, we have a solution
`F(x,y)=C` such that


`F_x=\mu M`

`F=\displaystyle\int(49y^3+45xy)x^(5/2)y^4\,\mathrm dx`

`F=10x^(9/2)y^5+14x^(7/2)y^7+f(y)`


`F_y=\mu N`

`50x^(9/2)y^4+98x^(7/2)y^6+f'(y)=98x^(7/2)y^2+50x^(9/2)y^4`

`f'(y)=0`

`\implies f(y)=C`

and so the general solution is


`F(x,y)=10x^(9/2)y^5+14x^(7/2)y^7=C`