Math 240 Home, Textbook Contents, Online Homework Home

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

### First Order Bernoulli Equations

Solve the following initial value problem \begin{align}\frac{dy}{dx} - 7y &= 3\sin(7x)y^{-3}\\ y(0) &= 2 \end{align} This is a Bernoulli equation. First we find the general solution following the paradigm.

1. We substitute $y = v^{1/(1-(-3)} = v^{1/4}$, so $dy/dx = (1/4)v^{-3/4}dv/dx$, and our equation becomes $$(1/4)v^{-3/4}\frac{dv}{dx} - 7v^{1/4} = 3\sin(7x)v^{-3/4}$$
2. Multiply by $4v^{3/4}$ to obtain a linear equation in the usual form. $$\frac{dv}{dx} - 28v = 12\sin(7x)$$
3. Solve the linear equation.

1. Find the integrating factor $$\mu(x) = \exp\left(\int -28 dx \right) = \exp(-28x)$$
2. Multiply through by the integrating factor $$\exp(-28x)\frac{dv}{dx} - 28\exp(-28x)v = 12\sin(7x)\exp(-28x)$$
3. Recognize the left-hand-side as $\displaystyle \frac{d}{dx}(\mu(x)v).$ $$\frac{d}{dx}(\exp(-28x)v) =12\sin(7x)\exp(-28x)$$
4. Integrate both sides. In this case you will need to integrate by parts twice and then solve for the unknown integral to evaluate the integral on the right (or you can use a table of integrals). $$\exp(-28x)v = (-(48/119)\sin(7x) - (12/119)\cos(7x))\exp(-28x) + C$$
5. Divide through by $\mu(x)$ to solve for $v.$ $$v = -(48/119)\sin(7x) - (12/119)\cos(7x) + C\exp(28x)$$

4. Back substitute for $y.$
5. $$y = (-(48/119)\sin(7x) - (12/119)\cos(7x) + C\exp(28x))^{1/4}$$
6. We check, but $y = 0$ is not a solution to this equation.

Now we plug in the initial values $x = 0$ and $y = 2$ and solve for $C = 1916/119,$ to obtain the solution to the initial value problem $$y = ( -(48/119)\sin(7x) - (12/119)\cos(7x) + (1916/119)\exp(28x) )^{1/4}$$ You may reload this page to generate additional examples.