Solve the three Bernoulli equations shown on the whiteboard.

I will solve the three Bernoulli equations shown on the whiteboard.

The general form of a Bernoulli equation is $\frac{dy}{dx} + P(x)y = Q(x)y^n$.

1) Solve $x \frac{dy}{dx} + y = x^2 y^2$

Step 1: Rewrite the equation in standard Bernoulli form.
Divide by $x$:

$$\frac{dy}{dx} + \frac{1}{x}y = xy^2$$

Here, $P(x) = \frac{1}{x}$, $Q(x) = x$, and $n=2$.

Step 2: Apply the substitution $u = y^{1-n}$.
For $n=2$, $u = y^{1-2} = y^{-1} = \frac{1}{y}$.
This implies $y = u^{-1}$.
Differentiate $y$ with respect to $x$: $\frac{dy}{dx} = -u^{-2} \frac{du}{dx}$.

Step 3: Substitute $y$ and $\frac{dy}{dx}$ into the equation.

$$-u^{-2} \frac{du}{dx} + \frac{1}{x}u^{-1} = x(u^{-1})^2$$ $$-\frac{1}{u^2} \frac{du}{dx} + \frac{1}{xu} = \frac{x}{u^2}$$

Step 4: Convert to a linear first-order differential equation.
Multiply by $-u^2$:

$$\frac{du}{dx} - \frac{u}{x} = -x$$

This is a linear equation of the form $\frac{du}{dx} + P_1(x)u = Q_1(x)$, with $P_1(x) = -\frac{1}{x}$ and $Q_1(x) = -x$.

Step 5: Find the integrating factor (IF).

$$IF = e^{\int P_1(x) dx} = e^{\int -\frac{1}{x} dx} = e^{-\ln|x|} = e^{\ln|x^{-1}|} = x^{-1} = \frac{1}{x}$$

Step 6: Multiply the linear equation by the integrating factor.

$$\frac{1}{x} \frac{du}{dx} - \frac{1}{x^2}u = -x \cdot \frac{1}{x}$$ $$\frac{d}{dx}\left(\frac{1}{x}u\right) = -1$$

Step 7: Integrate both sides with respect to $x$.

$$\int \frac{d}{dx}\left(\frac{1}{x}u\right) dx = \int -1 dx$$ $$\frac{1}{x}u = -x + C$$

Step 8: Solve for $u$ and substitute back $u = \frac{1}{y}$.

$$u = -x^2 + Cx$$ $$\frac{1}{y} = Cx - x^2$$

Step 9: Solve for $y$.

$$y = \frac{1}{Cx - x^2}$$

The final answer is $\boxed{y = \frac{1}{Cx - x^2}}$.

2) Solve $\frac{dy}{dx} + xy = xy^3$

Step 1: Identify the components of the Bernoulli equation.
The equation is already in the standard form $\frac{dy}{dx} + P(x)y = Q(x)y^n$.
Here, $P(x) = x$, $Q(x) = x$, and $n=3$.

Step 2: Apply the substitution $u = y^{1-n}$.
For $n=3$, $u = y^{1-3} = y^{-2} = \frac{1}{y^2}$.
This implies $y = u^{-1/2}$.
Differentiate $y$ with respect to $x$: $\frac{dy}{dx} = -\frac{1}{2}u^{-3/2} \frac{du}{dx}$.

Step 3: Substitute $y$ and $\frac{dy}{dx}$ into the equation.

$$-\frac{1}{2}u^{-3/2} \frac{du}{dx} + x u^{-1/2} = x(u^{-1/2})^3$$ $$-\frac{1}{2}u^{-3/2} \frac{du}{dx} + x u^{-1/2} = x u^{-3/2}$$

Step 4: Convert to a linear first-order differential equation.
Multiply by $-2u^{3/2}$:

$$\frac{du}{dx} - 2x u^{(3/2 - 1/2)} = -2x$$ $$\frac{du}{dx} - 2xu = -2x$$

This is a linear equation with $P_1(x) = -2x$ and $Q_1(x) = -2x$.

Step 5: Find the integrating factor (IF).

$$IF = e^{\int P_1(x) dx} = e^{\int -2x dx} = e^{-x^2}$$

Step 6: Multiply the linear equation by the integrating factor.

$$e^{-x^2} \frac{du}{dx} - 2x e^{-x^2} u = -2x e^{-x^2}$$ $$\frac{d}{dx}(e^{-x^2}u) = -2x e^{-x^2}$$

Step 7: Integrate both sides with respect to $x$.

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