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$ $$ \frac{dy}{dx} = \frac{5 - 3(y/x)}{-9 - (y/x)} $$
1. Substitute $ v=y/x,$ hence $ y=vx$ and by the product rule, $ dy/dx = v + x dv/dx$ $$ v + x \frac{dv}{dx} = \frac{5 - 3v}{-9 - v} $$
2. Solve this separable equation
   1. Separate the variables. $$ x \frac{dv}{dx} = \frac{5 - 3v}{-9 - v} - v = \frac{5 + 6v + v^2}{-9 - v} $$ $$ \frac{(-9 - v)dv}{5 + 6v + v^2} = \frac{dx}{x} $$
   2. Integrate both sides.

      We use partial fractions to rewrite $$ \frac{-9 - v}{5 + 6v + v^2} = \frac{1}{v + 5}- \frac{2}{v + 1} $$ We can then integrate both sides of the equation in substep i to get $$ \log|v + 5| - 2\log|v + 1| = \log|x| + C $$

   3. Solve for v

      We exponentiate both sides to get $$ \frac{(v + 5)^{}}{(v + 1)^{2}} = kx $$ Which we can also write as $$ (v + 5)^{} = kx(v + 1)^{2} $$

   4. Check for singular solutions.

      We divided by $5 + 6v + v^2$ which is 0 when $ v = -5$ and $ v = -1.$ 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 $$ (y/x + 5)^{} = kx(y/x + 1)^{2} $$ Multiplying both sides by $x^{}$ we can simplify this to $$ (y + 5x)^{} = k(y + x)^{2} $$ Finally, our singular solutions are $ y/x = -5$ and $ y/x = -1$, which simplify to $ y = -5x$ and $ y = -x.$

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