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### First Order Homogeneous Equations

#### Additional Examples

Solve the following initial value problem,
$$
\begin{align}
\frac{dy}{dx} &= \frac{20x + 3y}{-6x - y} \\
y(-1) &= 5
\end{align}
$$
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$ $$ \frac{dy}{dx} = \frac{20 + 3(y/x)}{-6 - (y/x)} $$
- Substitute $ v=y/x,$ hence $ y=vx$ and by the product rule, $ dy/dx = v + x dv/dx$ $$ v + x \frac{dv}{dx} = \frac{20 + 3v}{-6 - v} $$
- Solve this separable equation
- Separate the variables. $$ x \frac{dv}{dx} = \frac{20 + 3v}{-6 - v} - v = \frac{20 + 9v + v^2}{-6 - v} $$ $$ \frac{(-6 - v)dv}{20 + 9v + v^2} = \frac{dx}{x} $$
- Integrate both sides.
We use partial fractions to rewrite $$ \frac{-6 - v}{20 + 9v + v^2} = \frac{1}{v + 5}- \frac{2}{v + 4} $$ We can then integrate both sides of the equation in substep i to get $$ \log|v + 5| - 2\log|v + 4| = \log|x| + C $$

- Solve for v
We exponentiate both sides to get $$ \frac{(v + 5)^{}}{(v + 4)^{2}} = kx $$ Which we can also write as $$ (v + 5)^{} = kx(v + 4)^{2} $$

- Check for singular solutions.
We divided by $20 + 9v + v^2$ which is 0 when $ v = -5$ and $ v = -4.$ 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 $$ (y/x + 5)^{} = kx(y/x + 4)^{2} $$ Multiplying both sides by $x^{}$ we can simplify this to $$ (y + 5x)^{} = k(y + 4x)^{2} $$ Finally, our singular solutions are $ y/x = -5$ and $ y/x = -4$, which simplify to $ y = -5x$ and $ y = -4x.$

Now we plug in our initial value, $ y(-1) = 5.$ We get $ (0)^{} = k(1)^{2}$ and we solve this for $ k = 0,$ so the solution to the initial value problem is
$$
y = -5x
$$
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If you have any problems with this page, please contact bennett@math.ksu.edu.

©2010, 2014 Andrew G. Bennett