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

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

Now substitute

$v = \dfrac yx \implies y = xv \implies dy = x\,dv + v\,dx$

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.

$\dfrac{5v}{7+12v^2} \, dv = -\dfrac{dx}x$

Integrate both sides. On the left, substitute

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

$\displaystyle \int \frac{5v}{7+12v^2} \, dv = -\int \frac{dx}x$

$\displaystyle \dfrac5{24} \int \frac{dw}w = -\int \frac{dx}x$

$\dfrac5{24} \ln|w| = -\ln|x| + C$

Solve for $w$.

$\ln\left|w^{5/24}\right| = \ln\left|\dfrac1x\right| + C$

$\exp\left(\ln\left|w^{5/24}\right|\right) = \exp\left(\ln\left|\dfrac1x\right| + C\right)$

$w^{5/24} = \dfrac Cx$

Put this back in terms of $v$.

$(7+12v^2)^{5/24} = \dfrac Cx$

Put this back in terms of $y$.

$\left(7+12\dfrac{y^2}{x^2}\right)^{5/24} = \dfrac Cx$

Solve for $y$.

$7+12\dfrac{y^2}{x^2} = \dfrac C{x^{24/5}}$

$\dfrac{y^2}{x^2} = \dfrac C{x^{24/5}} - \dfrac7{12}$

$y^2= \dfrac C{x^{14/5}} - \dfrac{7x^2}{12}$

$y = \pm \sqrt{\dfrac C{x^{14/5}} - \dfrac{7x^2}{12}}$