# Differential Equations : Linear & Exact Equations

## Example Questions

### Example Question #1 : Linear & Exact Equations

Find the general solution of the given differential equation and determine if there are any transient terms in the general solution.

Explanation:

First, divide by  on both sides of the equation.

Identify the factor  term.

Integrate the factor.

Substitute this value back in and integrate the equation.

Now divide by  to get the general solution.

The transient term means a term that when the values get larger the term itself gets smaller. Therefore the transient term for this function is .

### Example Question #11 : Differential Equations

Find the solution for the following differential equation:

where .

Explanation:

This equation can be put into the form  as follows:

. Differential equations in this form can be solved by use of integrating factor. To solve, take  and solve for

Note, when using integrating factors, the +C constant is irrelevant as we only need one solution, not infinitely many. Thus, we have set C to 0.

Next, note that

Or more simply, . Integrating both sides using substitution of variables we find

Finally dividing by , we see

. Plugging in our initial condition,

So

And .

### Example Question #1 : Linear & Exact Equations

Consider the differential equation

Which of the terms in the differential equation make the equation nonlinear?

The    term makes the differential equation nonlinear.

The  term makes the differential equation nonlinear.

The  term makes the differential equation nonlinear

The  term makes the differential equation nonlinear

The    term makes the differential equation nonlinear.

Explanation:

The term  makes the differential equation nonlinear because a linear equation has the form of

### Example Question #4 : Linear & Exact Equations

Find the general solution of the given differential equation and determine if there are any transient terms in the general solution.

Explanation:

First, divide by  on both sides of the equation.

Identify the factor  term.

Integrate the factor.

Substitute this value back in and integrate the equation.

Now divide by  to get the general solution.

The transient term means a term that when the values get larger the term itself gets smaller. Therefore the transient term for this function is

### Example Question #1 : Linear & Exact Equations

Is the following differential equation exact?

If so, find the general solution.

No. The partial derivatives in the equation do not have the correct relationship.

Yes.

Yes.

No. The equation does not take the proper form.

Yes.

Yes.

Explanation:

For a differential equation to be exact, two things must be true. First, it must take the form . In our case, this is true, with  and . The second condition is that . Taking the partial derivatives, we find that  and . As these are equal, we have an exact equation.

Next we find a  such that  and . To do this, we can integrate  with respect to  or we can integrate  with respect to  Here, we choose arbitrarily to integrate .

We aren't quite done yet, because when taking a multivariate integral, the constant of integration can now be a function of y instead of just a constant. However, we know that , so taking the partial derivative, we find that  and thus that  and .

We now know that , and the point of finding psi was so that we could rewrite , and because the derivative of psi is 0, we know it must have been a constant. Thus, our final answer is

.

If you have an initial value, you can solve for c and have an implicit solution.

### Example Question #11 : Differential Equations

Is the following differential equation exact?  If so, find the general solution.

Yes.

No. The equation does not take the right form.

Yes.

No. The partial derivatives in the equation do not have the right relationship.

Yes.

No. The partial derivatives in the equation do not have the right relationship.

Explanation:

For a differential equation to be exact, two things must be true. First, it must take the form . In our case, this is true, with  and . The second condition is that . Taking the partial derivatives, we find that  and . As these are unequal, we do not have an exact equation.

### Example Question #1 : Linear & Exact Equations

Solve the Following Equation

Explanation:

Since this is in the form of a linear equation

we calculate the integration factor

Multiplying by  we get

Integrating

Plugging in the Initial Condition to solve for the Constant we get

Our solution is

### Example Question #1 : Linear & Exact Equations

Find the general solution of the differential equation

Explanation:

This is a Bernoulli Equation of the form

which requires a substitution

to transform it into a linear equation

Rearranging our equation gives us

Substituting

Solving the linear ODE gives us

Substituting in  and solving for

### Example Question #1 : Linear & Exact Equations

Solve the differential equation