Welcome to our study on Ordinary Differential Equations (ODEs)! In this article, we’ll dive into first-order ODEs, focusing on separable equations. We will explore separable equations, solve them, and provide practical examples to help you understand the concepts thoroughly. If you haven’t already, it’s a good idea to read the <introductory article> first, as it covers the foundational concepts that will make this section easier to follow.

First-order ODEs involve the first derivative of the unknown function with respect to the independent variable. As we learned in the introductory article, the general form of a first-order ODE is:

$\frac{dy}{dx} = f(x, y)\$

Here, $y \$ is the dependent variable, $x$ is the independent variable, and $f(x, y)$ is a function of $x$ and $y$ .

An ODE is said to be separable if the function $f(x,y)$ can be separated into the product of two functions, one in $x$ and the other in $y$ . A first-order ODE is called separable if it can be written as:

\frac{dy}{dx} = g(x)h(y)

This allows us to separate the variables:

\frac{1}{h(y)} \, dy = g(x) \, dx

Then integrate both sides:

\int \frac{1}{h(y)} \, dy = \int g(x) \, dx

1. Examples

Example 1:

Solve the equation:

\frac{dy}{dx} = y \cdot \cos(x)

Solution:

1. Separate the Variables:

- Rearrange the equation to separate the variables $y$ and $x$ :

$\frac{1}{y} \, dy = \cos(x) \, dx \$

2. Integrate Both Sides:

- Integrate both sides with respect to their respective variables here antiderivative of $\cos(x)\$ is $\sin(x)\$ :

$\int \frac{1}{y} \, dy = \int \cos(x) \, dx \$

$\ln|y| = \sin(x) + C_1 \$

3. Solve for $y(x)\$ :

- Exponentiate both sides to solve for $y\$ :

$y = e^{\sin(x) + C_1} = e^{\sin(x)} \cdot e^{C_1} \$

- Since $e^{C_1}\$ is a constant, we can write it as $C\$ :

$y = C \cdot e^{\sin(x)}\$

where $C\$ is a constant.

Note:

In this solution, the constant $C$ represents a different constant than the one in the equation $\ln|y| = \sin(x) + C_1$ . The constant $C_1\$ becomes $e^{C_1}\$ , and we rename it to $C\$ for simplicity.

Example 2:

Solve the equation:

\frac{dy}{dx} = y^2 \cdot \sin(x)

Solution:

1. Separate the variables:
$\frac{1}{y^2} \, dy = \sin(x) \, dx\$

2. Integrate both sides. Here antiderivative of $sin(x)\$ is $-\cos(x)\$ :
$\int \frac{1}{y^2} \, dy = \int \sin(x) \, dx\$

3. Solve the integrals:

$-\frac{1}{y} = -\cos(x) + C\$

where $C$ is the constant of integration.

4. Simplify and solve for $y$ :

$y = \frac{1}{\cos(x) - C}\$

Example 3:

Solve the equation:

\frac{dy}{dx} = \frac{y}{x}

Solution:

1. Separate the variables:

$\frac{1}{y} \, dy = \frac{1}{x} \, dx\$

2. Integrate both sides:
$\int \frac{1}{y} \, dy = \int \frac{1}{x} \, dx\$

3. Solve the integrals:

$ln|y| = \ln|x| + C\$

4. Simplify and solve for $y \$ :

$y = e^{\ln x + C} \$

3. Use the property of exponents $e^{\ln x + C} = e^{\ln x} \cdot e^C \$ :

$y = x \cdot e^C \$

4. Since $e^C \$ is just a constant, we can replace $e^C \$ with a new constant, say $K \$ :

$y = Kx \$

where $K \$ and $C$ are constants.

Can you try the examples below as questions and check the answer?

Example 4:

Example 4

Solve the initial value problem:

\frac{dy}{dx} = (1 + y^2) \cos(x)

y(0) = 0

Answer

Solution:

1. Separate the variables:

$\frac{1}{1 + y^2} \, dy = \cos(x) \, dx\$

2. Integrate both sides here antiderivative of $\frac{1}{1 + y^2}\$ is $\tan^{-1}(y)\$ :

$\int \frac{1}{1 + y^2} \, dy = \int \cos(x) \, dx\$

3. Solve the integrals:

$\tan^{-1}(y) = \sin(x) + C\$

4. Use the initial condition $y(0) = 0 \$ :

$\tan^{-1}(0) = \sin(0) + C \quad \Rightarrow \quad C = 0\$

5. So, the solution is:

$\tan^{-1}(y) = \sin(x)\$

$y = \tan(\sin(x))\$

Example 5:

Example 5

Solve the differential equation:

\frac{dv}{du} = v \cos(u)

Answer

Solution:

1. Separate the variables:

$\frac{1}{v} \, dv = \cos(u) \, du\$

2. Integrate both sides:

$\int \frac{1}{v} \, dv = \int \cos(u) \, du\$

3. Perform the integration:

$\ln|v| = \sin(u) + C\$

Where C is the constant of integration.

4. Solve for $v \$ :

Exponentiate both sides to eliminate the logarithm:

$v = e^{\sin(u) + C}\$

This can be written as:

$v(u) = Ke^{\sin(u)}\$

where $K = e^C \$ is a constant.

Example 6:

Example 6

Solve the differential equation:

\frac{dx}{dy} = x^2 - 1

Answer

Solution:

1. Separate the variables:

$\frac{1}{x^2 - 1} \, dx = dy\$

2. Factor and decompose the left-hand side:

$\frac{1}{(x-1)(x+1)} \, dx = dy\$

Using partial fractions:

$\frac{1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}\$

Solving for $A \$ and $B \$ :

$\frac{1}{(x-1)(x+1)} = \frac{1/2}{x-1} - \frac{1/2}{x+1}\$

Thus:

$\frac{1}{x^2 - 1} = \frac{1/2}{x-1} - \frac{1/2}{x+1}\$

3. Integrate both sides:

$\int \left( \frac{1/2}{x-1} - \frac{1/2}{x+1} \right) \, dx = \int dy\$

$\frac{1}{2} \left( \ln|x-1| - \ln|x+1| \right) = y + C\$

4. Simplify and solve for $x \$ :

Exponentiate both sides to get rid of the logarithms:

$\left| \frac{x-1}{x+1} \right| = e^{2(y + C)}\$

Let $K = e^{2C} \$ , then:

$\left| \frac{x-1}{x+1} \right| = Ke^{2y}\$

5. Solve for $x \$ :

Removing the absolute value and solving for $x \$ :

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

Solving for $x \$ :

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

Example 7:

Example 7

Solve the differential equation:

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

y(0) = 1

Answer

Solution:

1. Factor the right-hand side:

$\frac{dy}{dx} = y(x^2 - 1)\$

2. Separate the variables:

$\frac{1}{y} \, dy = (x^2 - 1) \, dx\$

3. Integrate both sides:

$\int \frac{1}{y} \, dy = \int (x^2 - 1) \, dx\$

The left-hand side integrates to:

$\ln|y| = \frac{x^3}{3} - x + C\$

4. Exponentiate both sides to solve for $y \$ :

$y(x) = e^{\frac{x^3}{3} - x + C}\$

5. Apply the initial condition $y(0) = 1 \$ :

$1 = e^{0 + 0 + C} \implies C = 0\$

So the solution is:

$y(x) = e^{\frac{x^3}{3} - x}\$

These problems and solutions illustrate how to approach differential equations involving different types of functions.

2. Practice Questions Set

Solve the following separable equations;

$\frac{dy}{dx} = 2x(1-y^2) \$
$\frac{dy}{dx} = \frac{xy}{x+1} \$
$2y dx + 3x^2 dy = 0 \$

Not sure about your answers? Don't worry, click the link below to check the video where the answers are explained.

Practice Questions Discussion

In this section, we explored separable equations.

Remember, understanding these methods takes time, and it’s perfectly okay to revisit the material until you feel confident in solving these types of equations.

In the next article, we will delve into more advanced techniques for solving first-order ODEs, such as Linear equations. Stay tuned, and happy solving!

Related Articles

3. Separable Equations

1. Examples

Example 4

Example 5

Example 6

Example 7

2. Practice Questions Set