First Order Differential Equations

First Order Differential Equations 1. First Order Linear Equations A first order linear differential equation has the following form:

The general solution is given by

where

called the integrating factor. If an initial condition is given, use it to find the constant C. Here are some practical steps to follow: 1. If the differential equation is given as

, rewrite it in the form

, where

2. Find the integrating factor .

3. Evaluate the integral 4. Write down the general solution

. 5. If you are given an IVP, use the initial condition to find the constant C. Example1: Find the particular solution of:

Solution: Let us use the steps: Step 1: There is no need for rewriting the differential equation. We have

Step 2: Integrating factor . Step 3: We have

. Step 4: The general solution is given by

. Step 5: In order to find the particular solution to the given IVP, we use the initial condition to find C. Indeed, we have .

Therefore the solution is . Note that you may not have to do the last step if you are asked to find the general solution (not an IVP). Example2: Find the solution to

Solution: First, we recognize that this is a linear equation. Indeed, we have

Therefore, the integrating factor is given by

.

Since

, we get

Hence, the general solution is given by the formula

We have

The details for this calculation involve the technique of integrating rational functions. We have

Hence, the only difficulty is in the integral

parts. We will differentiate t and integrate

. Here we will use integration by

. The details are left to the reader. We have

Therefore, we have

, which clearly implies

The general solution can also be rewritten as

Finally, the initial condition y(0) = 0.4 gives C = 0.4. Therefore, the solution to the IVP is

Example3: Solve the following initial value problem for t > 0

Answer: This is a linear equation. Let us follow these steps for solving such equations: 1. We have to divide by 2t

2. We get the integration factor u(t) by

3. The general solution is given by

. Since

Therefore, we have

4. The solution to the given initial value problem may be obtained by using the initial condition y(2)=4. We have ,

which gives

. Therefore, the solution is

Example4: Find the solution to

Answer: This is a linear equation. First we have to rewrite the equation with no function in front of y'. We get

, which may also be rewritten as

. Hence, the integrating factor is given by

Therefore, the general solution can be obtained as

Since we have

we get

The initial condition

implies ,

which gives C=-1. Therefore, the particular solution to the initial value problem is