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### First Order Homogeneous Equations

#### Additional Examples

Solve the following initial value problem,
$$
\begin{align}
\frac{dy}{dx} &= \frac{6y}{8x - y} \\
y(0) &= -6
\end{align}
$$
This is a homogeneous equation. First we find the general solution following the paradigm.

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©2010, 2014 Andrew G. Bennett

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- We divide the right hand side top and bottom by $ x$ to write the right hand side as a function of $ y/x$ $$ \frac{dy}{dx} = \frac{6(y/x)}{8 - (y/x)} $$
- Substitute $ v=y/x,$ hence $ y=vx$ and by the product rule, $ dy/dx = v + x dv/dx$ $$ v + x \frac{dv}{dx} = \frac{6v}{8 - v} $$
- Solve this separable equation
- Separate the variables. $$ x \frac{dv}{dx} = \frac{6v}{8 - v} - v = \frac{0 - 2v + v^2}{8 - v} $$ $$ \frac{(8 - v)dv}{0 - 2v + v^2} = \frac{dx}{x} $$
- Integrate both sides.
We use partial fractions to rewrite $$ \frac{8 - v}{0 - 2v + v^2} = \frac{3}{v - 2}- \frac{4}{v} $$ We can then integrate both sides of the equation in substep i to get $$ 3\log|v - 2| - 4\log|v| = \log|x| + C $$

- Solve for v
We exponentiate both sides to get $$ \frac{(v - 2)^{3}}{v^{4}} = kx $$ Which we can also write as $$ (v - 2)^{3} = kxv^{4} $$

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

- Back substitute.

$ v = y/x$ so our general solution becomes $$ (y/x - 2)^{3} = kx(y/x)^{4} $$ Multiplying both sides by $x^{3}$ we can simplify this to $$ (y - 2x)^{3} = ky^{4} $$ Finally, our singular solutions are $ y/x = 2$ and $ y/x = 0$, which simplify to $ y = 2x$ and $ y = 0.$

Now we plug in our initial value, $ y(0) = -6.$ We get $ (-6)^{3} = k(-6)^{4}$ and we solve this for $ k = -1/6,$ so the solution to the initial value problem is
$$
(y - 2x)^{3} = -(1/6)(y)^{4}
$$
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If you have any problems with this page, please contact bennett@math.ksu.edu.

©2010, 2014 Andrew G. Bennett