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### First Order Bernoulli Equations

Solve the following initial value problem \begin{align}\frac{dy}{dx} + y &= (-3x - 6)y^{4}\\ y(0) &= 4 \end{align} This is a Bernoulli equation. First we find the general solution following the paradigm.

1. We substitute $y = v^{1/(1-4)} = v^{-1/3}$, so $dy/dx = -(1/3)v^{-4/3}dv/dx$, and our equation becomes $$-(1/3)v^{-4/3}\frac{dv}{dx} + v^{-1/3} = (-3x - 6)v^{-4/3}$$
2. Multiply by $-3v^{4/3}$ to obtain a linear equation in the usual form. $$\frac{dv}{dx} - 3v = 9x + 18$$
3. Solve the linear equation.

1. Find the integrating factor $$\mu(x) = \exp\left(\int -3 dx \right) = \exp(-3x)$$
2. Multiply through by the integrating factor $$\exp(-3x)\frac{dv}{dx} - 3\exp(-3x)v = (9x + 18)\exp(-3x)$$
3. Recognize the left-hand-side as $\displaystyle \frac{d}{dx}(\mu(x)v).$ $$\frac{d}{dx}(\exp(-3x)v) =(9x + 18)\exp(-3x)$$
4. Integrate both sides. In this case you will need to integrate by parts to evaluate the integral on the right. $$\exp(-3x)v = (-3x - 7)\exp(-3x) + C$$
5. Divide through by $\mu(x)$ to solve for $v.$ $$v = -3x - 7+ C\exp(3x)$$

4. Back substitute for $y.$
5. $$y = (-3x - 7+ C\exp(3x))^{-1/3}$$
6. We check that $y = 0$ is indeed a singular solution.

Now we plug in the initial values $x = 0$ and $y = 4$ and solve for $C = 449/64,$ to obtain the solution to the initial value problem $$y = ( -3x - 7 + (449/64)\exp(3x) )^{-1/3}$$ You may reload this page to generate additional examples.