Separable differential equations 2

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For instance, on the planet Earth, man had always assumed that he was more intelligent than dolphins because he had achieved so much—the wheel, New York, wars and so on—whilst all the dolphins had ever done was muck about in the water having a good time. But conversely, the dolphins had always believed that they were far more intelligent than man —for precisely the same reasons, The Hitchhiker’s Guide to the Galaxy

Recall. Algebraic and differential equations.

An algebraic equation is a mathematical statement that declares or asserts the equality of two algebraic expressions. These expressions are constructed using:

1. Dependent and independent variables. Variables are symbols representing unknown quantities. Variables can be:

   Independent Variables: Variables that can be chosen freely.
   Dependent Variables: Variables that depend on the independent variables.

2. Constants. Fixed numerical values that do not change.

3. Algebraic operations. Operations such as addition, subtraction, multiplication, division, exponentiation, and root extraction.

An algebraic equation typically has the following form: Expression₁ = Expression₂, where the two expressions are set equal to each other.

Examples: y = 2x + 4 (simple linear equation, represents a straight line with a slope of 2 and a y-intercept of 4), ax² + bx + c = 0 (quadratic equation, it represents a parabola), $ln(xy + t) = z·sin(\sqrt{x})$.

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).

Separation of Variables

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.

Steps to Solve Separable Differential Equations

1. Check for any values of y that make g(y) = 0. These correspond to constant (equilibrium) solutions where $\frac{dy}{dx} = 0$.
2. Rewrite the equation to isolate y terms on one side and x terms on the other: $\frac{dy}{g(y)} = f(x)dx$
3. Integrate both sides of the equation: $\int \frac{dy}{g(y)} = \int f(x)dx$
4. Solve the resulting equation for y if possible to find the general solution.
5. If an initial condition y(x₀ = y₀) is provided, substitute x₀ and y₀ into the general solution and solve for the constant of integration C.

Solved exercises

• Solve the differential equation $\frac{dy}{dx} = e^x\sqrt{1-y^2}$ with initial conditions y(0) = 1 (a) and y(0) = 3 (b)

Solution: [1. Separate variables] $\frac{dy}{\sqrt{1-y^2}} = e^xdx ⇒[\text{2. Integrate Both Sides}] \int \frac{dx}{\sqrt{1-y^2}} = \int e^xdx ⇒ arcsin(y) = e^x + C ⇒[\text{3. Solve for y}] y = sin(e^x+c)$. This is the general solution to the differential equation.

Applying Initial Conditions. Apply y(0) = 1 (a). Substitute x = 0 and y = 1 into the general solution: into the previous equation, 1 = sin(1 + c), 1 + c = ^π⁄₂ + 2πn. For simplicity, choose n = 0, c = ^π⁄₂ -1. Therefore, our general solution is $y = sin(e^x+\frac{π}{2}-1)$.

Apply y(0) = 3 (b) into the previous equation, 3 = sin(1 + c), ⊥ The sine function has a range of [−1, 1]. The initial condition y(0) = 3 is not possible for the given differential equation.

• Solve the differential equation: $\frac{dy}{dx} = xy + 2x -2y -4$.

Step 1: Rearrange the Equation. Initially, the equation does not readily fit into standard forms like separable, linear, or exact equations. However, with some algebraic manipulation, we can transform it into a separable equation, which allows us to solve it by integrating both sides.

$\frac{dy}{dx} =[\text{Rearranging Terms}] xy + 2x -2y -4 = x(y+2) -2(y+2) = (x-2)(y+2)$.

$\frac{dy}{dx} = (x-2)(y+2) ↭[\text{2. Separate Variables}] \frac{dy}{y+2} = (x-2)dx ⇒[\text{ 3. Integrate Both Sides}] \int \frac{dy}{y+2} = \int (x-2)dx ⇒ln|y+2| = \frac{1}{2}x^2 -2x + C⇒[\text{4. Exponentiate Both Sides and solve for y}] |y + 2| = e^{\frac{1}{2}x^2 -2x + C} = e^Ce^{\frac{1}{2}x^2 -2x} ⇒ y +2 = ±Ke^{\frac{1}{2}x^2 -2x}$

However, since ±K is an arbitrary constant, it can be replaced with a new arbitrary constant C′, which can be any nonzero real number (positive or negative). Thus, we write: $y +2 = C’e^{\frac{1}{2}x^2 -2x} ⇒y = C’e^{\frac{1}{2}x^2 -2x} -2$. Since C’ is an arbitrary constant, it is customary to rename it back to C for simplicity: $y = Ce^{\frac{1}{2}x^2 -2x} -2$. This is the general solution to the differential equation.

Finding a Function with a Given Slope Condition

Let f(x) be a function such that the slope of the tangent to its curve at any point (x, y) is twice the slope of the ray from the origin to that point.

The slope of the tangent to the curve at point (x, y)is given by the derivative: $\frac{dy}{dx}$. The slope of the ray from the origin to the point (x, y): m_ray = $\frac{y}{x}$

The slope of the tangent is twice the slope of the ray from the origin: $\frac{dy}{dx}=2\frac{y}{x}$. This equation relates y, x, and their derivatives, forming a separable differential equation.

Solving the Differential Equation

Step 1. Separate Variables: $\frac{dy}{y} = \frac{2dx}{x}$

Step 2. Integrate Both Sides: $\int \frac{dy}{y} = \int \frac{2dx}{x}$

Step 3: Compute the Integrals: ln|y| = 2ln|x| + C.

Step 4: Exponentiate both sides to eliminate the natural logarithm and solve for y: $e^{ln|y|} = e^{2ln|x| + C} ⇒|y| = e^{2ln|x|}e^{C} ⇒ |y| = e^{ln(|x|^2)}e^{C} ⇒ |y| = Ax^2~ where~ A=e^c$ is a positive number.

Step 5: General solution: $y = ±Ax^2$. Let a = ±A be any real constant (positive or negative). The general solution is: y = ax².

Step 6 (Optional): Verification. To verify that y = ax² satisfies the original differential equation, compute:

1. $\frac{dy}{dx} = \frac{d}{dx}(ax^2) = 2ax$.
2. $2\frac{y}{x} = 2\frac{ax^2}{x} = 2ax$

Since both expressions are equal, y = ax² satisfies the differential equation. While the solution y = ax² is defined at x = 0, the differential equation $\frac{dy}{y} = \frac{2dx}{x}$ is undefined at x = 0 due to division by zero. This means the behavior of the solution at x = 0 cannot be determined from the differential equation alone, and x = 0 is a singular point.

Finding Orthogonal Trajectories to a Family of Parabolas

Calculate the family of curves perpendicular (orthogonal trajectories) to the family of parabolas defined by y = ax².

Orthogonal Trajectories are curves that intersect a given family of curves at right angles (90 degrees). Condition for Orthogonality. If two curves have slopes m₁ and m₂ at their point of intersection, they are orthogonal if: m₁·m₂ = 1.

Finding the Slope of the Parabolas. Given the family of parabolas y = ax².

Differentiating both sides with respect to x: $\frac{dy}{dx}=2ax$ =[Substituting a = $\frac{y}{x^2}$ (from the equation of the parabola)] = $2(\frac{y}{x^2})x = \frac{2y}{x}$. Thus, the slope of the tangent to the parabola at point (x, y) is: $m_{parabola} = \frac{2y}{x}$

Finding the Orthogonal Trajectories. Using the condition for orthogonality: m_parabola·m_orthogonal = -1 and solve for m_orthogonal: $m_{orthogonal} = -\frac{1}{m_{parabola}} = -\frac{x}{2y}$

Write the Differential Equation. The differential equation representing the orthogonal trajectories is: $\frac{dy}{dx} = -\frac{x}{2y}$.

Solving the Differential Equation

[1. Separate variables] 2ydy = -xdx ⇒[2. Taking antiderivates] $\int 2ydy = \int -xdx ↭[\text{3. Compute the integrals}] y^2 = \frac{-x^2}{2} + C ⇒ \frac{x^2}{2}+y^2 = C$.

[4. Interpretation]. The equation $\frac{x^2}{2}+y^2 = C$ represents a family of ellipses centered at the origin. These ellipses are the orthogonal trajectories to the family of parabolas y = ax² (Figure 1.b.). The parabolas open upwards or downwards, depending on the value of a. The ellipses intersect these parabolas at right angles.

Consider that the slope at any point (x, y) of an orthogonal curve is the negative reciprocal $-\frac{1}{m}$ of the slope at the point (x, y) of the parabola passing through that point.

Applying Newton’s Law of Cooling

A freshly baked cake is taken out of the oven where it had been baking at a temperature of approximately 350°F. The ambient temperature of the kitchen is 75°F. After just 5 minutes, the temperature of the cake drops to 340°F. How much additional time will it take for the cake to cool down to 300°F?

When an object is removed from a heat source and placed in a cooler environment, it begins to cool down. The rate at which it cools is described by Newton’s Law of Cooling, which states that the rate of change of the temperature of an object (the rate at which an object cools) is proportional to the difference between its own temperature and the ambient temperature: $\frac{dT}{dt} = -k(T -T_{ambient})$ where:

• T(t) is the temperature of the object (cake) at time t.
• T_ambient is the ambient temperature, (75°F).
• k is a positive constant of proportionality (cooling rate),
• $\frac{dT}{dt}$ is the rate of change of temperature with respect to time.

Given T_ambient = 75°F, the differential equation becomes: $\frac{dT}{dt} = -k(T-75)$. This is a first-order linear differential equation with initial conditions T(0) = 350 (temperature of the cake when it comes out of the oven. This is the initial temperature).

Solving the Differential Equation: $\frac{dT}{dt} = -k(T-75) ↭[\text{1. Separable variables}] \frac{dT}{t-75} = -kdt ⇒[\text{2. Integrate both sides}] \int \frac{dT}{t-75} = -k\int dt ↭[\text{3. Compute the integrals}] ln|T-75| + C_1 = -kt + C_2 ↭[\text{4. Solve for T by exponentiating both sides,}~ C = C_2 - C_1] e^{ln|T-75|} = e^{-kt + C}↭ |T -75| = C’e^{-kt} ⇒[\text{Since T − 75 > 0 (the cake is hotter than the ambient temperature), we can drop the absolute value:}] T -75 = C’e^{-kt}$ where $C’ = e^C$.

Apply Initial Conditions to Find C’: T(0) = 350F, plug in t = 0 and T = 350 into the equation: $T = 75 + C’e^{-kt} ↭ 350 = 75 + C’ ⇒ C = 275.$ Therefore, the general solution to the initial-value problem is $T = 75 + 275e^{-kt}$, expressing the temperature T as a function of time t.

Use Additional Data to Find k. We are given that after 5 minutes, the temperature is 340°F ⇒T(5) = $340 = 75 + 275e^{-5k}↭ 265 = 275e^{-5k}↭ e^{-5k} = \frac{265}{275} = \frac{53}{55}↭ ln(e^{-5k}) = ln(\frac{53}{55})↭ -5k = -0.037 ↭ k = \frac{0.037}{5}≈ 0.007408$.

Substitute k back into the general solution: $300 = 75 + 275e^{−0.007408t}$

How much longer will we have to wait until the temperature of the cake reaches 300°F? $300 = 75 + 275e^{−0.007408t} ↭ 225 = 275e^{−0.007408t} ↭ e^{−0.007408t} = \frac{225}{275} = \frac{9}{11} ↭ ln(e^{−0.007408t}) = ln(\frac{9}{11}) ⇒ −0.007408t = ln(\frac{9}{11}) ↭ t = \frac{ln(\frac{9}{11})}{−0.007408}≈ \frac{−0.2007}{−0.007408} = 27.0884$ minutes.

We have found that the total time from when the cake was removed from the oven until it reaches 300°F is approximately 27.0884 minutes. Since 5 minutes have already passed, the additional time we need to wait approximately is t_additional = 27.0884 - 5 = 22.0884 minutes. It will take approximately 22.0884 minutes more for the cake to cool down to 300°F.

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