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$ $$ \frac{dy}{dx} = \frac{35 + 11(y/x)}{-1 - (y/x)} $$
1. Substitute $ v=y/x,$ hence $ y=vx$ and by the product rule, $ dy/dx = v + x dv/dx$ $$ v + x \frac{dv}{dx} = \frac{35 + 11v}{-1 - v} $$
2. Solve this separable equation
   1. Separate the variables. $$ x \frac{dv}{dx} = \frac{35 + 11v}{-1 - v} - v = \frac{35 + 12v + v^2}{-1 - v} $$ $$ \frac{(-1 - v)dv}{35 + 12v + v^2} = \frac{dx}{x} $$
   2. Integrate both sides.

      We use partial fractions to rewrite $$ \frac{-1 - v}{35 + 12v + v^2} = \frac{2}{v + 5}- \frac{3}{v + 7} $$ We can then integrate both sides of the equation in substep i to get $$ 2\log|v + 5| - 3\log|v + 7| = \log|x| + C $$

   3. Solve for v

      We exponentiate both sides to get $$ \frac{(v + 5)^{2}}{(v + 7)^{3}} = kx $$ Which we can also write as $$ (v + 5)^{2} = kx(v + 7)^{3} $$

   4. Check for singular solutions.

      We divided by $35 + 12v + v^2$ which is 0 when $ v = -5$ and $ v = -7.$ We check these are both solutions (though $ v = -5$ is already in the general solution when $ k = 0$).

3. Back substitute.

$ v = y/x$ so our general solution becomes $$ (y/x + 5)^{2} = kx(y/x + 7)^{3} $$ Multiplying both sides by $x^{2}$ we can simplify this to $$ (y + 5x)^{2} = k(y + 7x)^{3} $$ Finally, our singular solutions are $ y/x = -5$ and $ y/x = -7$, which simplify to $ y = -5x$ and $ y = -7x.$

Now we plug in our initial value, $ y(0) = -3.$ We get $ (-3)^{2} = k(-3)^{3}$ and we solve this for $ k = -1/3,$ so the solution to the initial value problem is $$ (y + 5x)^{2} = -(1/3)(y + 7x)^{3} $$ You may reload this page to generate additional examples.