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$ $$ \frac{dy}{dx} = \frac{5 + 13(y/x)}{7 - (y/x)} $$
1. Substitute $ v=y/x,$ hence $ y=vx$ and by the product rule, $ dy/dx = v + x dv/dx$ $$ v + x \frac{dv}{dx} = \frac{5 + 13v}{7 - v} $$
2. Solve this separable equation
   1. Separate the variables. $$ x \frac{dv}{dx} = \frac{5 + 13v}{7 - v} - v = \frac{5 + 6v + v^2}{7 - v} $$ $$ \frac{(7 - v)dv}{5 + 6v + v^2} = \frac{dx}{x} $$
   2. Integrate both sides.

      We use partial fractions to rewrite $$ \frac{7 - v}{5 + 6v + v^2} = \frac{2}{v + 1}- \frac{3}{v + 5} $$ We can then integrate both sides of the equation in substep i to get $$ 2\log|v + 1| - 3\log|v + 5| = \log|x| + C $$

   3. Solve for v

      We exponentiate both sides to get $$ \frac{(v + 1)^{2}}{(v + 5)^{3}} = kx $$ Which we can also write as $$ (v + 1)^{2} = kx(v + 5)^{3} $$

   4. Check for singular solutions.

      We divided by $5 + 6v + v^2$ which is 0 when $ v = -1$ and $ v = -5.$ We check these are both solutions (though $ v = -1$ is already in the general solution when $ k = 0$).

3. Back substitute.

$ v = y/x$ so our general solution becomes $$ (y/x + 1)^{2} = kx(y/x + 5)^{3} $$ Multiplying both sides by $x^{2}$ we can simplify this to $$ (y + x)^{2} = k(y + 5x)^{3} $$ Finally, our singular solutions are $ y/x = -1$ and $ y/x = -5$, which simplify to $ y = -x$ and $ y = -5x.$

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