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First Order Homogeneous Equations

Additional Examples

Solve the following initial value problem, $$ \begin{align} \frac{dy}{dx} &= \frac{32x + 12y}{-y} \\ y(-1) &= 9 \end{align} $$ This is a homogeneous equation. First we find the general solution following the paradigm.

  1. We divide the right hand side top and bottom by $ x$ to write the right hand side as a function of $ y/x$ $$ \frac{dy}{dx} = \frac{32 + 12(y/x)}{-(y/x)} $$
  2. Substitute $ v=y/x,$ hence $ y=vx$ and by the product rule, $ dy/dx = v + x dv/dx$ $$ v + x \frac{dv}{dx} = \frac{32 + 12v}{-v} $$
  3. Solve this separable equation

    1. Separate the variables. $$ x \frac{dv}{dx} = \frac{32 + 12v}{-v} - v = \frac{32 + 12v + v^2}{-v} $$ $$ \frac{-vdv}{32 + 12v + v^2} = \frac{dx}{x} $$
    2. Integrate both sides.

      We use partial fractions to rewrite $$ \frac{-v}{32 + 12v + v^2} = \frac{1}{v + 4}- \frac{2}{v + 8} $$ We can then integrate both sides of the equation in substep i to get $$ \log|v + 4| - 2\log|v + 8| = \log|x| + C $$

    3. Solve for v

      We exponentiate both sides to get $$ \frac{(v + 4)^{}}{(v + 8)^{2}} = kx $$ Which we can also write as $$ (v + 4)^{} = kx(v + 8)^{2} $$

    4. Check for singular solutions.

      We divided by $32 + 12v + v^2$ which is 0 when $ v = -4$ and $ v = -8.$ We check these are both solutions (though $ v = -4$ is already in the general solution when $ k = 0$).

  4. Back substitute.
  5. $ v = y/x$ so our general solution becomes $$ (y/x + 4)^{} = kx(y/x + 8)^{2} $$ Multiplying both sides by $x^{}$ we can simplify this to $$ (y + 4x)^{} = k(y + 8x)^{2} $$ Finally, our singular solutions are $ y/x = -4$ and $ y/x = -8$, which simplify to $ y = -4x$ and $ y = -8x.$

Now we plug in our initial value, $ y(-1) = 9.$ We get $ (5)^{} = k(1)^{2}$ and we solve this for $ k = 5,$ so the solution to the initial value problem is $$ (y + 4x) = 5(y + 8x)^{2} $$ You may reload this page to generate additional examples.


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©2010, 2014 Andrew G. Bennett