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### First Order Homogeneous Equations

#### Additional Examples

Solve the following initial value problem,
$$
\begin{align}
\frac{dy}{dx} &= \frac{12x - 2y}{5x - y} \\
y(1) &= 6
\end{align}
$$
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$ $$ \frac{dy}{dx} = \frac{12 - 2(y/x)}{5 - (y/x)} $$
- Substitute $ v=y/x,$ hence $ y=vx$ and by the product rule, $ dy/dx = v + x dv/dx$ $$ v + x \frac{dv}{dx} = \frac{12 - 2v}{5 - v} $$
- Solve this separable equation
- Separate the variables. $$ x \frac{dv}{dx} = \frac{12 - 2v}{5 - v} - v = \frac{12 - 7v + v^2}{5 - v} $$ $$ \frac{(5 - v)dv}{12 - 7v + v^2} = \frac{dx}{x} $$
- Integrate both sides.
We use partial fractions to rewrite $$ \frac{5 - v}{12 - 7v + v^2} = \frac{1}{v - 4}- \frac{2}{v - 3} $$ We can then integrate both sides of the equation in substep i to get $$ \log|v - 4| - 2\log|v - 3| = \log|x| + C $$

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

- Check for singular solutions.
We divided by $12 - 7v + v^2$ which is 0 when $ v = 4$ and $ v = 3.$ We check these are both solutions (though $ v = 4$ is already in the general solution when $ k = 0$).

- Back substitute.

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

Now we plug in our initial value, $ y(1) = 6.$ We get $ (2)^{} = k(3)^{2}$ and we solve this for $ k = 2/9,$ so the solution to the initial value problem is
$$
(y - 4x) = 2/9(y - 3x)^{2}
$$
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If you have any problems with this page, please contact bennett@math.ksu.edu.

©2010, 2014 Andrew G. Bennett