Math 240 Home,
Textbook Contents,
Online Homework Home

**Warning: MathJax
requires JavaScript to process the mathematics on this page.**

If your browser supports JavaScript, be sure it is enabled.**
****
**

### First Order Homogeneous Equations

#### Additional Examples

Solve the following initial value problem,
$$
\begin{align}
\frac{dy}{dx} &= \frac{45x - 21y}{-7x - y} \\
y(1) &= 12
\end{align}
$$
This is a homogeneous equation. First we find the general solution following the paradigm.

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

©2010, 2014 Andrew G. Bennett

If your browser supports JavaScript, be sure it is enabled.

- We divide the right hand side top and bottom by $ x$ to write the right hand side as a function of $ y/x$ $$ \frac{dy}{dx} = \frac{45 - 21(y/x)}{-7 - (y/x)} $$
- Substitute $ v=y/x,$ hence $ y=vx$ and by the product rule, $ dy/dx = v + x dv/dx$ $$ v + x \frac{dv}{dx} = \frac{45 - 21v}{-7 - v} $$
- Solve this separable equation
- Separate the variables. $$ x \frac{dv}{dx} = \frac{45 - 21v}{-7 - v} - v = \frac{45 - 14v + v^2}{-7 - v} $$ $$ \frac{(-7 - v)dv}{45 - 14v + v^2} = \frac{dx}{x} $$
- Integrate both sides.
We use partial fractions to rewrite $$ \frac{-7 - v}{45 - 14v + v^2} = \frac{3}{v - 5}- \frac{4}{v - 9} $$ We can then integrate both sides of the equation in substep i to get $$ 3\log|v - 5| - 4\log|v - 9| = \log|x| + C $$

- Solve for v
We exponentiate both sides to get $$ \frac{(v - 5)^{3}}{(v - 9)^{4}} = kx $$ Which we can also write as $$ (v - 5)^{3} = kx(v - 9)^{4} $$

- Check for singular solutions.
We divided by $45 - 14v + v^2$ which is 0 when $ v = 5$ and $ v = 9.$ 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 $$ (y/x - 5)^{3} = kx(y/x - 9)^{4} $$ Multiplying both sides by $x^{3}$ we can simplify this to $$ (y - 5x)^{3} = k(y - 9x)^{4} $$ Finally, our singular solutions are $ y/x = 5$ and $ y/x = 9$, which simplify to $ y = 5x$ and $ y = 9x.$

Now we plug in our initial value, $ y(1) = 12.$ We get $ (7)^{3} = k(3)^{4}$ and we solve this for $ k = 343/81,$ so the solution to the initial value problem is
$$
(y - 5x)^{3} = 343/81(y - 9x)^{4}
$$
*You may reload this page to generate additional examples.*

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

©2010, 2014 Andrew G. Bennett