Welcome to our study on Ordinary Differential Equations (ODEs)! In this article, we’ll dive into first-order ODEs, focusing on direct integration. We will explore direct integration to 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 previous 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 $y$ .

Direct Integration is one of the simplest methods to solve first-order ODEs. It is applicable when the ODE can be written in the form: $\frac{dy}{dx} = f(x)\$ . In this case, the equation can be directly integrated with respect to $x$ to find $y(x) \$ since $f(x)$ consists only of $x$ terms.

1. Examples

Example 1:

Find the general solution of the equation:

\frac{dy}{dx} = 3x^2

Solution:

Integrate both sides with respect to $x$ :

$y = \int 3x^2 \, dx\$

$y = x^3 + C\$

where $C$ is the constant of integration.

Example 2:

Find the general solution of the equation:

\frac{dy}{dx} = \sin(x)

Solution:

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

$y = \int \sin(x) \, dx\$

$y = -\cos(x) + C\$

where $C$ is the constant of integration.

Example 3:

Find the general solution of the equation:

\frac{dy}{dx} = e^{2x} + 4x^3

Solution:

Integrate both sides with respect to $x$ :

$y = \int (e^{2x} + 4x^3) \, dx\$

Solve the integrals separately. Here, we use the fact that the anti-derivative of $e^{nx}\$ is $\frac{e^{nx}}{n}\$ :

$y = \frac{1}{2}e^{2x} + x^4 + C\$

where $C$ is the constant of integration.

$\frac{dy}{dx}$ can be also represented as $y'$ which we call as $y$ prime.

Example 4:

Example 4

Find the solution to the equation:

y' = \frac{1}{1+x^2}

y(1) = 0

Answer

To find the general solution, integrate both sides with respect to $x$ :

$y = \int (8x^3 + 10x) \, dx\$

This gives:
$y = \frac{8x^4}{4} + \frac{10x^2}{2} + C = 2x^4 + 5x^2 + C\$

where $C$ is the constant of integration.

$\frac{dy}{dx}$ can be also represented as $\dot{y}\$ which we call $y$ dot.

Example 5:

Example 5

Find the general solution of the equation:

\dot{y} = \frac{1}{t^2}

Answer

To find the general solution, integrate both sides with respect to $t$ :

$y = \int \frac{1}{t^2} \, dt\$

This gives:

$y = \int t^{-2} \, dt = \frac{t^{-1}}{-1} + C = -\frac{1}{t} + C\$

where $C$ is the constant of integration.

Example 6:

Example 6

Find the general solution of the equation:

\frac{dw}{dt} = \sec^2(t)

Answer

To find the general solution, integrate both sides with respect to the antiderivative of $\sec^2 (t)$ is $\tan(t)$ :

$w = \int \sec^2(t) \, dt\$

This gives:

$w = \tan(t) + C\$

where $C$ is the constant of integration.

Example 7:

Example 7

Find the general solution of the equation:

\dot{x} = 6e^{-3t}

Answer

To find the general solution, integrate both sides with respect to $t$ :

$x = \int 6e^{-3t} \, dt\$

This gives:

$x = 6 \cdot \frac{e^{-3t}}{-3} + C = -2e^{-3t} + C\$

where $C$ is the constant of integration.

Example 8:

Example 8

Find the solution of the equation:

\frac{dq}{dt}=4t^3 - 6t^2

q(0)=2

Answer

First, find the general solution by integrating both sides with respect to $t$ :

$q(t) = \int (4t^3 - 6t^2) \, dt\$

This gives:

$q(t) = \frac{4t^4}{4} - \frac{6t^3}{3} + C = t^4 - 2t^3 + C\$

Now, use the initial condition $q(0) = 2 \$ to find $C$ :

$2 = 0^4 - 2(0)^3 + C \implies C = 2\$

Thus, the solution is:

$q(t) = t^4 - 2t^3 + 2\$

Example 9:

Example 9

Find the solution of the equation:

\frac{dv}{dy} = 3u^2 - \sin(u)

v(0)=5

Answer

First, find the general solution by integrating both sides with respect to $u$

here antiderivative of $sin(u)\$ is $-\cos(u)\$ :

$v(u) = \int (3u^2 - \sin(u)) \, du\$

This gives:

$v(u) = \int 3u^2 \, du - \int \sin(u) \, du\$

$v(u) = u^3 + \cos(u) + C\$

Now, use the initial condition $v(0) = 5 \$ to find $C$ :

$5 = 0^3 + \cos(0) + C \implies 5 = 1 + C \implies C = 4\$

Thus, the solution is:

$v(u) = u^3 + \cos(u) + 4\$

Although this article focuses on first-order ordinary differential equations (ODEs), the method of solving a second-order ODE by treating it as a first-order equation through substitution can be a useful technique.

Example 1:

Given the differential equation:

$\frac{d^2y}{dt^2} = e^t\$

Solution:

1. Substitution:

- Let $\frac{dy}{dt} = w(t)\$ .

- Then, $\frac{d^2y}{dt^2} = \frac{dw}{dt}\$ .

The equation becomes:

$\frac{dw}{dt} = e^t \$

2. Integrate to Find $w(t)\$ :

- Integrate with respect to t to find the first derivative $w(t)\$ :

$w(t) = \int e^t \, dt + C_1 \$

- The integral of $e^t\$ with respect to t is $e^t\$ :

$w(t) = e^t + C_1 \$

- Since $w(t) = \frac{dy}{dt}\$ , we can rewrite this as:

$\frac{dy}{dt} = e^t + C_1\$

3. Integrate to Find $y(t)\$ :

- Integrate again to find $y(t)\$ :

$y(t) = \int (e^t + C_1) \, dt + C_2 \$

- Integrating each term, we get the solution as:

$y(t) = e^t + C_1t + C_2 \$

$C_1 , C_2 \$ are constants.

Can you give it a try on your own and check the answers later?

Let's solve the second-order differential equation step-by-step, similar to the approach used in the previous example.

Example 2:

Example 2

Solve the differential equation:

$\frac{d^2y}{dt^2} = t^2\$

Answer

Solution:

1. Substitution:

- Let $\frac{dy}{dt} = w(t)\$ .

- Then, $\frac{d^2y}{dt^2} = \frac{dw}{dt}\$ .

The equation becomes:

$\frac{dw}{dt} = t^2 \$

2. Integrate to Find $w(t)\$ :

- Integrate with respect to $t\$ to find the first derivative $w(t)\$ :

$w(t) = \int t^2 \, dt + C_1 \$

- The integral of $t^2\$ with respect to $t\$ is $\frac{t^3}{3}\$ :

$w(t) = \frac{t^3}{3} + C_1\$

- Since $w(t) = \frac{dy}{dt}\$ , we can rewrite this as:

$\frac{dy}{dt} = \frac{t^3}{3} + C_1 \$

3. Integrate to Find $y(t)\$ :

- Integrate again to find $y(t)\$ :

$y(t) = \int \left(\frac{t^3}{3} + C_1\right) \, dt + C_2 \$

- Integrating each term we get the solution:

$y(t) = \frac{t^4}{12} + C_1t + C_2 \$

$C_1 , C_2 \$ are constants.

Example 3:

Example 3

Solve the differential equation:

$\frac{d^2y}{dt^2} = \ln(t+1)\$

Answer

Solution:

1. Substitution:

- Let $\frac{dy}{dt} = w(t)\$ .

- Then, $\frac{d^2y}{dt^2} = \frac{dw}{dt}\$ .

The equation becomes:

$\frac{dw}{dt} = \ln(t+1) \$

2. Integrate to Find $w(t)\$ :

- Integrate with respect to t to find the first derivative $w(t)\$ :

$w(t) = \int \ln(t+1) \, dt + C_1 \$

To integrate $\ln(t+1)\$ , we'll use integration by parts.

- Let $u = \ln(t+1) \ and \ dv = dt \$ .

- Then, $du = \frac{1}{t+1} \, dt \ and \ v = t .$

Using integration by parts:

$\int \ln(t+1) \, dt = t \ln(t+1) - \int \frac{t}{t+1} \, dt \$

To simplify $\int \frac{t}{t+1} \, dt\$ , use the substitution $u = t + 1 \$ , hence $du = dt \$ :

$\int \frac{t}{t+1} \, dt = \int \left(1 - \frac{1}{t+1}\right) \, dt = t - \ln(t+1) \$

Putting everything together:

$\int \ln(t+1) \, dt = t \ln(t+1) - (t - \ln(t+1)) = (t + 1) \ln(t+1) - t \$

Therefore:

$w(t) = (t + 1) \ln(t+1) - t + C_1 \$

Since $w(t) = \frac{dy}{dt}\$ , we can rewrite this as:

$\frac{dy}{dt} = (t + 1) \ln(t+1) - t + C_1 \$

3. Integrate to Find \y(t)\:

- Integrate again to find \y(t)\:

$y(t) = \int \left((t + 1) \ln(t+1) - t + C_1\right) \, dt + C_2 \$

Now, integrate each term:

- To integrate $(t + 1) \ln(t+1)\$ :

$\int (t + 1) \ln(t+1) \, dt = \frac{(t + 1)^2 \ln(t+1) - (t + 1)^2 + 2(t + 1)}{2} \$

- Integrate $-t\$ :

$\int -t \, dt = -\frac{t^2}{2} \$

- Integrate $C_1\$ :

$\int C_1 \, dt = C_1t \$

Combining everything we get the general solution:

$y(t) = \frac{(t + 1)^2 \ln(t+1) - (t + 1)^2 + 2(t + 1)}{2} - \frac{t^2}{2} + C_1t + C_2\ \ C_1 , C_2 \$ are constants.

"Practicing all these questions will help you build a solid understanding. Remember, practice makes perfect, so keep working on these problems to master the concepts!"

2. Practice Questions Set

Solve the following direct integration questions:

1. $y' = 8x^3 + 10x\$

2. $y' = \frac{1}{t^2}\, , t(0) = 1 \$

3. $x' = 6e^{-3t}, \quad x(0) = 1 \$

4. $y'' = 2t \$

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

Practice Questions Discussion

So, we focused on solving first-order differential equations using direct integration. This method worked well when we could easily integrate both sides of the equation with respect to their respective variables. However, not all differential equations are as straightforward. Now, we'll explore a new method called separation of variables.

Related Articles

2. Direct Integration

1. Examples

Example 4

Example 5

Example 6

Example 7

Example 8

Example 9

Example 2

Example 3

2. Practice Questions Set