Kansas State University

0. We divide the right hand side top and bottom by $ x$ to write the right hand side as a function of $ y/x$ $$ \frac{dy}{dx} = \frac{3 - 5(y/x)}{-9 - (y/x)} $$
1. Substitute $ v=y/x,$ hence $ y=vx$ and by the product rule, $ dy/dx = v + x dv/dx$ $$ v + x \frac{dv}{dx} = \frac{3 - 5v}{-9 - v} $$
2. Solve this separable equation
   1. Separate the variables. $$ x \frac{dv}{dx} = \frac{3 - 5v}{-9 - v} - v = \frac{3 + 4v + v^2}{-9 - v} $$ $$ \frac{(-9 - v)dv}{3 + 4v + v^2} = \frac{dx}{x} $$
   2. Integrate both sides.

      We use partial fractions to rewrite $$ \frac{-9 - v}{3 + 4v + v^2} = \frac{3}{v + 3}- \frac{4}{v + 1} $$ We can then integrate both sides of the equation in substep i to get $$ 3\log|v + 3| - 4\log|v + 1| = \log|x| + C $$

   3. Solve for v

      We exponentiate both sides to get $$ \frac{(v + 3)^{3}}{(v + 1)^{4}} = kx $$ Which we can also write as $$ (v + 3)^{3} = kx(v + 1)^{4} $$

   4. Check for singular solutions.

      We divided by $3 + 4v + v^2$ which is 0 when $ v = -3$ and $ v = -1.$ We check these are both solutions (though $ v = -3$ is already in the general solution when $ k = 0$).

3. Back substitute.

$ v = y/x$ so our general solution becomes $$ (y/x + 3)^{3} = kx(y/x + 1)^{4} $$ Multiplying both sides by $x^{3}$ we can simplify this to $$ (y + 3x)^{3} = k(y + x)^{4} $$ Finally, our singular solutions are $ y/x = -3$ and $ y/x = -1$, which simplify to $ y = -3x$ and $ y = -x.$

Now we plug in our initial value, $ y(0) = 8.$ We get $ (8)^{3} = k(8)^{4}$ and we solve this for $ k = 1/8,$ so the solution to the initial value problem is $$ (y + 3x)^{3} = (1/8)(y + x)^{4} $$ You may reload this page to generate additional examples.