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First Order Bernoulli Equations

Solve the following initial value problem \begin{align}\frac{dy}{dx} - 3y &= (8x + 9)y^{-2}\\ y(0) &= 1 \end{align} This is a Bernoulli equation. First we find the general solution following the paradigm.

1. We substitute $y = v^{1/(1-(-2)} = v^{1/3}$, so $dy/dx = (1/3)v^{-2/3}dv/dx$, and our equation becomes $$(1/3)v^{-2/3}\frac{dv}{dx} - 3v^{1/3} = (8x + 9)v^{-2/3}$$
2. Multiply by $3v^{2/3}$ to obtain a linear equation in the usual form. $$\frac{dv}{dx} - 9v = 24x + 27$$
3. Solve the linear equation.

1. Find the integrating factor $$\mu(x) = \exp\left(\int -9 dx \right) = \exp(-9x)$$
2. Multiply through by the integrating factor $$\exp(-9x)\frac{dv}{dx} - 9\exp(-9x)v = (24x + 27)\exp(-9x)$$
3. Recognize the left-hand-side as $\displaystyle \frac{d}{dx}(\mu(x)v).$ $$\frac{d}{dx}(\exp(-9x)v) =(24x + 27)\exp(-9x)$$
4. Integrate both sides. In this case you will need to integrate by parts to evaluate the integral on the right. $$\exp(-9x)v = (-(8/3)x - 89/27)\exp(-9x) + C$$
5. Divide through by $\mu(x)$ to solve for $v.$ $$v = -(8/3)x - 89/27+ C\exp(9x)$$

4. Back substitute for $y.$
5. $$y = (-(8/3)x - 89/27+ C\exp(9x))^{1/3}$$
6. We check, but $y = 0$ is not a solution to this equation.

Now we plug in the initial values $x = 0$ and $y = 1$ and solve for $C = 116/27,$ to obtain the solution to the initial value problem $$y = ( -(8/3)x - 89/27 + (116/27)\exp(9x) )^{1/3}$$ You may reload this page to generate additional examples.