Question

7(x^2 y^2)dx 5xydyUse the method for solving homogeneous equations to solve the following differential equation.

Answer 1

I'm guessing the equation should read something like

`7(x^2+y^2) \,dx + 5xy \, dy = 0`

or possibly with minus signs in place of +.

Multiply both sides by
`\frac1{x^2}` to get

`7\left(1 + (y^2)/(x^2)\right) \, dx + \frac{5y}x \, dy = 0`

Now substitute

`v = \frac yx \implies y = xv \implies dy = x\,dv + v\,dx`

to transform the equation to

`7(1+v^2) \, dx + 5v (x\,dv + v\,dx) = 0`

which simplifies to

`(7 + 12v^2) \, dx + 5xv\,dv = 0`

The ODE is now separable.

`(5v)/(7+12v^2) \, dv = -\frac{dx}x`

Integrate both sides. On the left, substitute

`w = 7+12v^2 \implies dw = 24v\, dv`

`\displaystyle \int (5v)/(7+12v^2) \, dv = -\int \frac{dx}x`

`\displaystyle \frac5{24} \int \frac{dw}w = -\int \frac{dx}x`

`\frac5{24} \ln|w| = -\ln|x| + C`

Solve for
`w`.

`\ln\left|w^(5/24)\right| = \ln\left|\frac1x\right| + C`

`\exp\left(\ln\left|w^(5/24)\right|\right) = \exp\left(\ln\left|\frac1x\right| + C\right)`

`w^(5/24) = \frac Cx`

Put this back in terms of
`v`.

`(7+12v^2)^(5/24) = \frac Cx`

Put this back in terms of
`y`.

`\left(7+12(y^2)/(x^2)\right)^(5/24) = \frac Cx`

Solve for
`y`.

`7+12(y^2)/(x^2) = \frac C{x^(24/5)}`

`(y^2)/(x^2) = \frac C{x^(24/5)} - \frac7{12}`

`y^2= \frac C{x^(14/5)} - (7x^2)/(12)`

`y = \pm \sqrt{\frac C{x^(14/5)} - (7x^2)/(12)}`