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### First Order Homogeneous Equations

#### Additional Examples

Solve the following initial value problem,
$$
\begin{align}
\frac{dy}{dx} &= \frac{-x + 7y}{7x - y} \\
y(0) &= 5
\end{align}
$$
This is a homogeneous equation. First we find the general solution following the paradigm.

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©2010 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$ $$ \frac{dy}{dx} = \frac{-1 + 7(y/x)}{7 - (y/x)} $$
- Substitute $ v=y/x,$ hence $ y=vx$ and by the product rule, $ dy/dx = v + x dv/dx$ $$ v + x \frac{dv}{dx} = \frac{-1 + 7v}{7 - v} $$
- Solve this separable equation
- Separate the variables. $$ x \frac{dv}{dx} = \frac{-1 + 7v}{7 - v} - v = \frac{-1 + v^2}{7 - v} $$ $$ \frac{(7 - v)dv}{-1 + v^2} = \frac{dx}{x} $$
- Integrate both sides.
We use partial fractions to rewrite $$ \frac{7 - v}{-1 + v^2} = \frac{3}{v - 1}- \frac{4}{v + 1} $$ We can then integrate both sides of the equation in substep i to get $$ 3\log|v - 1| - 4\log|v + 1| = \log|x| + C $$

- Solve for v
We exponentiate both sides to get $$ \frac{(v - 1)^{3}}{(v + 1)^{4}} = kx $$ Which we can also write as $$ (v - 1)^{3} = kx(v + 1)^{4} $$

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

- Back substitute.

$ v = y/x$ so our general solution becomes $$ (y/x - 1)^{3} = kx(y/x + 1)^{4} $$ Multiplying both sides by $x^{3}$ we can simplify this to $$ (y - x)^{3} = k(y + x)^{4} $$ Finally, our singular solutions are $ y/x = 1$ and $ y/x = -1$, which simplify to $ y = x$ and $ y = -x.$

Now we plug in our initial value, $ y(0) = 5.$ We get $ (5)^{3} = k(5)^{4}$ and we solve this for $ k = 1/5,$ so the solution to the initial value problem is
$$
(y - x)^{3} = (1/5)(y + x)^{4}
$$
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If you have any problems with this page, please contact bennett@math.ksu.edu.

©2010 Andrew G. Bennett