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### First Order Homogeneous Equations

#### Additional Examples

Solve the following initial value problem,
$$
\begin{align}
\frac{dy}{dx} &= \frac{10x + 4y}{11x - y} \\
y(1) &= 2
\end{align}
$$
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$ $$ \frac{dy}{dx} = \frac{10 + 4(y/x)}{11 - (y/x)} $$
- Substitute $ v=y/x,$ hence $ y=vx$ and by the product rule, $ dy/dx = v + x dv/dx$ $$ v + x \frac{dv}{dx} = \frac{10 + 4v}{11 - v} $$
- Solve this separable equation
- Separate the variables. $$ x \frac{dv}{dx} = \frac{10 + 4v}{11 - v} - v = \frac{10 - 7v + v^2}{11 - v} $$ $$ \frac{(11 - v)dv}{10 - 7v + v^2} = \frac{dx}{x} $$
- Integrate both sides.
We use partial fractions to rewrite $$ \frac{11 - v}{10 - 7v + v^2} = \frac{2}{v - 5}- \frac{3}{v - 2} $$ We can then integrate both sides of the equation in substep i to get $$ 2\log|v - 5| - 3\log|v - 2| = \log|x| + C $$

- Solve for v
We exponentiate both sides to get $$ \frac{(v - 5)^{2}}{(v - 2)^{3}} = kx $$ Which we can also write as $$ (v - 5)^{2} = kx(v - 2)^{3} $$

- Check for singular solutions.
We divided by $10 - 7v + v^2$ which is 0 when $ v = 5$ and $ v = 2.$ We check these are both solutions (though $ v = 5$ is already in the general solution when $ k = 0$).

- Back substitute.

$ v = y/x$ so our general solution becomes $$ (y/x - 5)^{2} = kx(y/x - 2)^{3} $$ Multiplying both sides by $x^{2}$ we can simplify this to $$ (y - 5x)^{2} = k(y - 2x)^{3} $$ Finally, our singular solutions are $ y/x = 5$ and $ y/x = 2$, which simplify to $ y = 5x$ and $ y = 2x.$

Now we plug in our initial value, $ y(1) = 2.$ When we plug this value in, we get the equation $9 = 0,$ which indicates that we can't satisfy the initial condition using our general solution. This suggests we consider the singular solution, and we recognize that this condition is satisfied by the singular solution, $$ y = 2x$$ so this is the solution to the initial value problem.

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If you have any problems with this page, please contact bennett@math.ksu.edu.

©2010, 2014 Andrew G. Bennett