The differential equations in parts a, d, and e appear to be of the Bernoulli type. Let's go step by step to solve each one.

Problem a:

$\frac{dy}{dx} - y = -y^2$

This is a Bernoulli differential equation of the form: $\frac{dy}{dx} + P(x)y = Q(x)y^n$ where $n = 2$, $P(x) = -1$, and $Q(x) = -1$.

Step 1: Divide by $y^2$ to transform the equation:

$y^{-2} \frac{dy}{dx} - y^{-1} = -1$ Let $v = y^{-1}$, then $\frac{dv}{dx} = -y^{-2} \frac{dy}{dx}$, so the equation becomes: $\frac{dv}{dx} + v = 1$

Step 2: Solve the linear differential equation:

This is a linear equation. The integrating factor is $e^{\int 1 dx} = e^x$. Multiplying through by the integrating factor: $e^x \frac{dv}{dx} + e^x v = e^x$ This simplifies to: $\frac{d}{dx}(e^x v) = e^x$ Integrating both sides: $e^x v = e^x + C$ Dividing by $e^x$: $v = 1 + Ce^{-x}$ Since $v = y^{-1}$, we get: $y = \frac{1}{1 + Ce^{-x}}$

Problem d:

$\frac{dy}{dx} - y = xy^2$

This is also a Bernoulli equation, where $P(x) = -1$, $Q(x) = x$, and $n = 2$.

Step 1: Divide by $y^2$:

$y^{-2} \frac{dy}{dx} - y^{-1} = x$ Let $v = y^{-1}$, so that $\frac{dv}{dx} = -y^{-2} \frac{dy}{dx}$, and the equation becomes: $\frac{dv}{dx} + v = x$

Step 2: Solve the linear equation:

The integrating factor is $e^{\int 1 dx} = e^x$. Multiplying through by the integrating factor: $e^x \frac{dv}{dx} + e^x v = xe^x$ This simplifies to: $\frac{d}{dx}(e^x v) = xe^x$ We integrate both sides using integration by parts: $e^x v = (x - 1)e^x + C$ Dividing by $e^x$: $v = x - 1 + Ce^{-x}$ Since $v = y^{-1}$, we get: $y = \frac{1}{x - 1 + Ce^{-x}}$

Problem e:

$x \frac{dy}{dx} + y + x^2 y^2 e^x = 0$

Again, this is a Bernoulli equation, where $P(x) = \frac{1}{x}$, $Q(x) = x e^x$, and $n = 2$.

Step 1: Divide by $y^2$:

$\frac{1}{y^2} x \frac{dy}{dx} + \frac{1}{y} = -x^2 e^x$ Let $v = y^{-1}$, and $\frac{dv}{dx} = -y^{-2} \frac{dy}{dx}$. The equation becomes: $-x \frac{dv}{dx} + v = -x^2 e^x$ Rearranging: $x \frac{dv}{dx} - v = x^2 e^x$

Step 2: Solve the linear equation:

The integrating factor is $e^{-\int \frac{1}{x} dx} = \frac{1}{x}$. Multiplying through by the integrating factor: $\frac{1}{x} x \frac{dv}{dx} - \frac{1}{x} v = e^x$ Simplifying: $\frac{d}{dx}(v) = e^x$ Integrating both sides: $v = e^x + C$ Since $v = y^{-1}$, we get: $y = \frac{1}{e^x + C}$

Summary of Solutions:

• (a) $y = \frac{1}{1 + Ce^{-x}}$
• (d) $y = \frac{1}{x - 1 + Ce^{-x}}$
• (e) $y = \frac{1}{e^x + C}$

Would you like further clarification or a more detailed breakdown of any of these steps?

5 Related Questions:

1. How do you recognize a Bernoulli differential equation?
2. What are the typical steps for solving a linear first-order differential equation?
3. How does the integrating factor method work in differential equations?
4. What is the role of substitution in solving Bernoulli equations?
5. Can Bernoulli equations always be reduced to linear equations?

Tip:

When solving Bernoulli equations, always look for substitutions that transform the non-linear terms into linear ones, simplifying the equation significantly.