Math 240 Home, Textbook Contents, Online Homework Home

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

### First Order Homogeneous Equations

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.

1. We divide the right hand side top and bottom by $x$ to write the right hand side as a function of $y/x$ $$\frac{dy}{dx} = \frac{-2 + 11(y/x)}{10 - (y/x)}$$
2. Substitute $v=y/x,$ hence $y=vx$ and by the product rule, $dy/dx = v + x dv/dx$ $$v + x \frac{dv}{dx} = \frac{-2 + 11v}{10 - v}$$
3. Solve this separable equation

1. Separate the variables. $$x \frac{dv}{dx} = \frac{-2 + 11v}{10 - v} - v = \frac{-2 + v + v^2}{10 - v}$$ $$\frac{(10 - v)dv}{-2 + v + v^2} = \frac{dx}{x}$$
2. 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 $$3\log|v - 1| - 4\log|v + 2| = \log|x| + C$$

3. Solve for v

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

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 $k = 0$).

4. Back substitute.
5. $v = y/x$ 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 $$(y - x)^{3} = -32(y + 2x)^{4}$$ You may reload this page to generate additional examples.