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### First Order Homogeneous Equations

#### Additional Examples

Solve the following initial value problem,
$$
\begin{align}
\frac{dy}{dx} &= \frac{-2x + 11y}{10x - y} \\
y(2) &= -6
\end{align}
$$
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$ $$ \frac{dy}{dx} = \frac{-2 + 11(y/x)}{10 - (y/x)} $$
- Substitute $ v=y/x,$ hence $ y=vx$ and by the product rule, $ dy/dx = v + x dv/dx$ $$ v + x \frac{dv}{dx} = \frac{-2 + 11v}{10 - v} $$
- Solve this separable equation
- Separate the variables. $$ x \frac{dv}{dx} = \frac{-2 + 11v}{10 - v} - v = \frac{-2 + v + v^2}{10 - v} $$ $$ \frac{(10 - v)dv}{-2 + v + v^2} = \frac{dx}{x} $$
- Integrate both sides.
We use partial fractions to rewrite $$ \frac{10 - v}{-2 + v + v^2} = \frac{3}{v - 1}- \frac{4}{v + 2} $$ We can then integrate both sides of the equation in substep i to get $$ 3\log|v - 1| - 4\log|v + 2| = \log|x| + C $$

- Solve for v
We exponentiate both sides to get $$ \frac{(v - 1)^{3}}{(v + 2)^{4}} = kx $$ Which we can also write as $$ (v - 1)^{3} = kx(v + 2)^{4} $$

- Check for singular solutions.
We divided by $-2 + v + v^2$ which is 0 when $ v = 1$ and $ v = -2.$ 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 $$ (y/x - 1)^{3} = kx(y/x + 2)^{4} $$ Multiplying both sides by $x^{3}$ we can simplify this to $$ (y - x)^{3} = k(y + 2x)^{4} $$ Finally, our singular solutions are $ y/x = 1$ and $ y/x = -2$, which simplify to $ y = x$ and $ y = -2x.$

Now we plug in our initial value, $ y(2) = -6.$ We get $ (-8)^{3} = k(-2)^{4}$ and we solve this for $ k = -32,$ so the solution to the initial value problem is
$$
(y - x)^{3} = -32(y + 2x)^{4}
$$
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If you have any problems with this page, please contact bennett@math.ksu.edu.

©2010, 2014 Andrew G. Bennett