I have yet to see any problem, however complicated, which, when looked at in the right way did not become still more complicated, Paul Anderson.

Definition. A differential equation is an equation that involves one or more functions and their derivatives. It relates the function itself (dependent variable), its derivatives with respect to one or more independent variables, and the independent variables themselves, e.g., $\frac{dy}{dx} = 3x +5y, y’ + y = 4xcos(2x), \frac{dy}{dx} = x^2y+y, etc.$

Key Components (e.g., $\frac{dy}{dx} = 3x +5y$):

**Dependent variables**:*Variable(s) (y) that depend on one or more other variables*. The function y (or y(x)) we are trying to solve for.**Independent variables**: Variable(s) (x) upon which the dependent variables depend.**Derivatives**: Expressions like $\frac{dy}{dx}$ or y’ that represent the rate at which the dependent variable (y) change with respect to the independent variable (x).

Separable differential equations are a special class of differential equations that can be manipulated algebraically to separate the variables, allowing us to integrate and find solutions.

A separable differential equation is any differential equation that can be written in the form $\frac{dy}{dx}=f(x)g(y)$. This form allows us to separate the variables x and y on opposite sides of the equation, facilitating integration.

**Check for any values of y that make g(y) = 0.**These correspond to constant (equilibrium) solutions where $\frac{dy}{dx} = 0$.**Rewrite the equation to isolate y terms on one side and x terms**on the other: $\frac{dy}{g(y)} = f(x)dx$**Integrate both sides**of the equation: $\int \frac{dy}{g(y)} = \int f(x)dx$**Solve the resulting equation for y**if possible to find the general solution.- If an initial condition y(x
_{0}= y_{0}) is provided, substitute x_{0}and y_{0}into the general solution and solve for the constant of integration C.

Definition. A Bernoulli equation is a type of first-order nonlinear differential equation of the form: y' + A(x)y = B(x)y^{r} where:

- A(x) and B(x) are continuous functions of x.
- y is the dependent variable.
- y′ is the derivative of y with respect to x.
- r is a real number, and r ≠ 0, 1.

**Special cases**:

- If r = 0, the equation simplifies to y’ + A(x)y=B(x), which is a linear first-order ODE.
- If r = 1, the equation simplifies to y′ +(A(x)−B(x))y=0, which is also a linear first-order ODE.

When r ≠ 0, 1, **the equation is nonlinear, and it requires a special technique to solve**, involving a substitution to transform it into a linear equation.

Step 1: **Write the Equation in Standard Form**. Ensure the equation is in the standard Bernoulli form: y’ + A(x)y = B(x)y^{r}

Step 2. **Solve the Associated Homogeneous Linear Equation**

Consider the associated homogeneous linear equation: y’ + A(x)y = 0. This is a linear first-order ODE and can be solved using separation of variables:

**Rearrange**: $y’ = -A(x)y$**Separate variables**: $\frac{dy}{y} = -A(x)dx$.**Integrate both sides**: $\int \frac{dy}{y} = \int -A(x)dx ⇒ ln|y| = \int -A(x)dx + C_1$**Taking exponentials on both sides and solve for y**: $|y| = e^{\int -A(x)dx + C_1} = C_2e^{\int -A(x)dx}⇒ y = ±C_2e^{\int -A(x)dx} = Ce^{\int -A(x)dx}$ where C = ±C_{2}.

We denote the solution of the homogeneous equation as y_{0}: $y_0 = Ce^{\int -A(x)dx}$. For simplicity, we often denote $y_0 = e^{\int -A(x)dx}$ that satisfies $y_0’ +A(x)y’_0 = 0$

Step 3: **Use the Substitution y = y _{0}z**

Substitute y = y_{0}z into the original Bernoulli equation and use the product rule. Compute
y′ using the product rule: y’ = y_{0}‘z + y_{0}z’, we get: y_{0}‘z + y_{0}z’ +A(x)y_{0}z = B(x)y_{0}^{r}z^{r}.

Step 4. **Simplify the Equation.**

y’_{0}z + A(x)y_{0}z = z(y_{0}’ + A(x)y_{0}) =[y_{0} is the solution of the homogeneous equation y_{0}’ + A(x)y_{0} = 0] z·0 = 0, hence y_{0}z’ = B(x)y_{0}^{r}z^{r} ⇒[Divide both side by y_{0}, by assumption r ≠ 0, 1] z’ = B(x)y_{0}^{r-1}z^{r}. Now, the equation for z is separable.

Step 5. **Separate variables**

$\frac{dz}{z^r} = B(x)y_0^{r-1}dx$, that we can integrate, then solve for z, and finally backtrack the substitution y = y_{0}z to find y.

- y’ - y = xy
^{2}

Step 1. The given equation is already in standard form, y’ + A(x)y = B(x)y^{r} where A(x) = −1
B(x) = x, and r = 2

Step 2: Solve the Homogeneous Equation: y’ -y = 0 ⇒[Rearrange] y’ = y ⇒[Separate variables] $\frac{dy}{y} = dx ⇒[\text{Integrate both sides}] \int \frac{dy}{y} = \int 1·dx ⇒ ln|y| = x + C ⇒[\text{Taking exponentials}] |y| = e^{x+C} = C_1e^x$. Denote the solution as y_{0}, $y_0 = e^x$

Step 3: Substitute y = y_{0}z = e^{x}z ⇒[Taking differentials] y’ = z’e^{x} + ze^{x}. Substitute y = e^{x}z and y’ = z’e^{x} + ze^{x} into the original equation:

z’e^{x} + ze^{x} -e^{x}z = x(z^{2}e^{2x})⇒[Simplify] z’e^{x} = xz^{2}e^{2x} ⇒[Multiplying both side by e^{-x}] z’ = xz^{2}e^{x} ⇒[Separate Variables] $\frac{dz}{z^2} = xe^xdx ⇒ [\text{Integrate Both Sides}] \int \frac{dz}{z^2} = \int xe^xdx ⇒ \frac{-1}{z} =[\text{Require integration by parts [*]}] xe^x-e^x+c ⇒[\text{Solve for z}] z = \frac{-1}{xe^x-e^x+c} ⇒[\text{Back-Substitute to Find y}] y = e^x·z = \frac{z}{e^{-x}} = \frac{-1}{e^{-x}(xe^x-e^x+c)} = \frac{-1}{x-1+ce^{-x}} = \frac{1}{(1-x)+Ce^{-x}}$

[*] Using the integration by parts formula ∫udv = uv −∫vdu, u = x⇒ du = dx, dv = e^{x}dx ⇒v = e^{x}: $\int xe^xdx = xe^x -\int e^xdx = xe^x -e^x + c.$

- $y’ + \frac{y}{x} = x^3y^3$

To solve the differential equation $y’ + \frac{y}{x} = x^3y^3$ we recognize it as a Bernoulli equation of the form: y′ + P(x)y = Q(x)y^{n} where P(x) = ^{1}⁄_{x}, Q(x) = x^{3} and n = 3.

The given equation is already in standard form. Consider the associated homogeneous equation: $y’ + \frac{y}{x} = 0 ⇒[\text{Separate variables:}] y’ = \frac{-y}{x} ⇒\frac{dy}{y} = \frac{-dx}{x} ⇒[\text{Integrate both sides}] \int \frac{dy}{y} = \int \frac{-dx}{x} ⇒ ln|y| = -ln|x| + C$

We are going to assume that whatever given conditions, we will always get y(x_{0}) = y_{0} > 0, x_{0} > 0. This assumption will let us get rid of the absolute values of the variables x and y.

$ln(y) = -ln(x) + C ⇒ ln(y) = ln(\frac{1}{x}) + C⇒[\text{Taking exponentials}] y = e^{ln(\frac{1}{x}) + C} = e^C\frac{1}{x}$

For simplicity, we often denote: $y_0 = \frac{1}{x}$

Substitute y = y_{0}z = $\frac{z}{x}$ into the original Bernoulli equation: $\frac{z’}{x}-\frac{z}{x^2}+\frac{z}{x^2} = x^3\frac{z^3}{x^3} ⇒[\text{Simplify the terms}] \frac{z’}{x} = z^3 ⇒[\text{Separate variables}] \frac{dz}{z^3} = xdx ⇒[\text{Integrate both sides}] \int \frac{dz}{z^3} = \int xdx ⇒ \frac{-1}{2z^2} = \frac{x^2}{2} + C ⇒[\text{Multiply by 2}] \frac{-1}{z^2} = x^2 + 2C ⇒ z^2 = \frac{1}{-2C-x^2}⇒[z^2 ≥ 0, \text{the denominator must be negative}] z = \frac{1}{\sqrt{-2C-x^2}} ⇒[\text{Back-Substitute z = yx}] y = \frac{z}{x} = \frac{1}{\sqrt{-2Cx^2-x^4}} = \frac{1}{x\sqrt{K-x^2}}$ where K = -2C is a constant determined by initial conditions.

- Solve $\frac{dy}{dx} = \sqrt{y}-y$ and y(0) = 9.

Step 1: **Rewrite the Equation in Standard Form**

The given differential equation is: $\frac{dy}{dx} = \sqrt{y}-y$. We can rearrange it to bring all terms involving y to one side: $\frac{dy}{dx} + y = \sqrt{y}$.

Step 2: **Recognize the Bernoulli Equation**

A Bernoulli equation is a first-order differential equation of the form: y’ + P(x)y = Q(x)y^{r} where P(x) = 1, Q(x) = 1, and r is any real number, r = ^{1}⁄_{2}.

Step 3: **Make the Appropriate Substitution**

Let v = $y^{1-r} = y^{1-\frac{1}{2}} = y^\frac{1}{2} ⇒ y = v^2$. Differentiate both sides with respect to x: $\frac{dy}{dx} = 2v\frac{dv}{dx}$

Step 4: **Substitute Back into the Original Equation**

Our equation is now: $2v\frac{dv}{dx}+v^2 = v↭[\text{Let’s divide by 2v}] \frac{dv}{dx}+ \frac{1}{2}v = \frac{1}{2}$. This is an ordinary linear first order differential equation of the form $\frac{dv}{dx}+Pv = Q$ where P(x) = 1/2 and Q(x) = 1/2.

Step 5: **Find the Integrating Factor**: $μ = e^{\int P(x)dx} =e^{\int \frac{1}{2}dx} = e^{\frac{1}{2}x}$

Step 6. **Multiply both sides of the equation by the integrating factor**: $e^{\frac{1}{2}x}\frac{dv}{dx} + e^{\frac{1}{2}x}\frac{1}{2}v = \frac{1}{2}e^{\frac{1}{2}x} ↭ \frac{d}{dx}(e^{\frac{1}{2}x}·v) = \frac{1}{2}e^{\frac{1}{2}x} ⇒[\text{Step 7. Integrating both sides}] e^{\frac{1}{2}x}·v = \int \frac{1}{2}e^{\frac{1}{2}x}dx ↭ e^{\frac{1}{2}x}·v = e^{\frac{1}{2}·x} + C ↭ [\text{Step 8. Divide by } e^{\frac{1}{2}·x} \text{and solve for v}] v = 1 + Ce^{\frac{-1}{2}x}$

Step 8. Recall $v = y^{\frac{1}{2}}$. $\sqrt{y} = 1 + Ce^{\frac{-1}{2}x}$

Step 9. Apply Initial conditions y(0) = 9, $\sqrt{9} = 1 + Ce^{\frac{-1}{2}0}↭ 3 = 1 + C ⇒ C = 2$. Then, $\sqrt{y} = 1 + 2e^{\frac{-1}{2}x}↭ y = (1 + 2e^{\frac{-1}{2}x})^2$ ∎

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