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$ $$ \frac{dy}{dx} = \frac{6 - 4(y/x)}{1 - (y/x)} $$
1. Substitute $ v=y/x,$ hence $ y=vx$ and by the product rule, $ dy/dx = v + x dv/dx$ $$ v + x \frac{dv}{dx} = \frac{6 - 4v}{1 - v} $$
2. Solve this separable equation
   1. Separate the variables. $$ x \frac{dv}{dx} = \frac{6 - 4v}{1 - v} - v = \frac{6 - 5v + v^2}{1 - v} $$ $$ \frac{(1 - v)dv}{6 - 5v + v^2} = \frac{dx}{x} $$
   2. Integrate both sides.

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

   3. Solve for v

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

   4. Check for singular solutions.

      We divided by $6 - 5v + v^2$ which is 0 when $ v = 2$ and $ v = 3.$ We check these are both solutions (though $ v = 2$ is already in the general solution when $ k = 0$).

3. Back substitute.

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

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