Let's solve the problem step-by-step. The task has two parts:

1. Solve the Bernoulli equation: $\frac{dy}{dx} + \frac{y}{2x} = \frac{x}{y^3}$ with the initial condition $y(1) = 2$.

2. Find an integrating factor for the linear differential equation: $\frac{dy}{dx} + \left(\frac{2x+1}{x}\right)y = e^{-2x}.$

Part 1: Solve the Bernoulli Equation

The general form of a Bernoulli equation is: $\frac{dy}{dx} + P(x)y = Q(x)y^n.$ Here, $n = -3$, $P(x) = \frac{1}{2x}$, and $Q(x) = x$.

Step 1: Divide through by $y^3$ to make it linear in $v = y^{-2}$.

Multiply both sides of the equation by $y^{-3}$: $y^{-3} \frac{dy}{dx} + \frac{y^{-2}}{2x} = x.$ Let $v = y^{-2}$, so $\frac{dv}{dx} = -2y^{-3} \frac{dy}{dx}$.

Rewriting: $-\frac{1}{2} \frac{dv}{dx} + \frac{v}{2x} = x.$

Multiply through by $-2$: $\frac{dv}{dx} - \frac{v}{x} = -2x.$

Step 2: Solve the linear equation for $v(x)$.

This is a first-order linear equation. The integrating factor is: $\mu(x) = e^{\int -\frac{1}{x} dx} = e^{-\ln|x|} = \frac{1}{x}.$

Multiply through by the integrating factor $\mu(x) = \frac{1}{x}$: $\frac{1}{x} \frac{dv}{dx} - \frac{v}{x^2} = -2.$ $\frac{d}{dx} \left( \frac{v}{x} \right) = -2.$

Integrate both sides: \frac{v}{x} = -2x + C, \quad \text{where C is a constant of integration.} $v = -2x^2 + Cx.$

Step 3: Back-substitute $v = y^{-2}$.

$y^{-2} = -2x^2 + Cx.$

Take the reciprocal and square root: $y = \pm \frac{1}{\sqrt{-2x^2 + Cx}}.$

Step 4: Use the initial condition $y(1) = 2$ to find $C$.

When $x = 1$, $y = 2$: $2^{-2} = -2(1)^2 + C(1).$ $\frac{1}{4} = -2 + C.$ $C = \frac{9}{4}.$

So the solution is: $y = \pm \frac{1}{\sqrt{-2x^2 + \frac{9}{4}x}}.$

Part 2: Find the Integrating Factor

For the equation: $\frac{dy}{dx} + \left(\frac{2x+1}{x}\right)y = e^{-2x},$ the integrating factor is: $\mu(x) = e^{\int \frac{2x+1}{x} dx}.$

Split the integral: $\int \frac{2x+1}{x} dx = \int \frac{2x}{x} dx + \int \frac{1}{x} dx = \int 2 dx + \int \frac{1}{x} dx.$ $\int \frac{2x+1}{x} dx = 2x + \ln|x|.$

Thus: $\mu(x) = e^{2x + \ln|x|} = e^{2x} \cdot x.$

The integrating factor is: $\mu(x) = xe^{2x}.$

Final Results:

1. The solution to the Bernoulli equation is: $y = \pm \frac{1}{\sqrt{-2x^2 + \frac{9}{4}x}}.$

2. The integrating factor for the linear equation is: $\mu(x) = xe^{2x}.$

Let me know if you'd like further clarification or additional steps!

Follow-up Questions:

1. How do you verify solutions for Bernoulli's equation using substitution?
2. What are the key differences between linear and Bernoulli differential equations?
3. Can you solve the linear equation after finding the integrating factor?
4. How is the initial condition used to find constants in the solution?
5. Why is the integrating factor $\mu(x)$ critical for solving linear differential equations?

Tip:

Always simplify and rewrite differential equations to identify their type (linear, separable, or Bernoulli) before solving.