### All Differential Equations Resources

## Example Questions

### Example Question #6 : First Order Differential Equations

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

**Possible Answers:**

**Correct answer:**

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

Find the solution for the following differential equation:

where .

**Possible Answers:**

**Correct answer:**

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?

**Possible Answers:**

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

**Correct answer:**

The term makes the differential equation nonlinear.

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

### Example Question #3 : Linear & Exact Equations

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

**Possible Answers:**

**Correct answer:**

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.

**Possible Answers:**

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

Yes.

No. The equation does not take the proper form.

Yes.

Yes.

**Correct answer:**

Yes.

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 #1 : Linear & Exact Equations

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

**Possible Answers:**

Yes.

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

Yes.

No. The equation does not take the right form.

Yes.

**Correct answer:**

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

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

**Possible Answers:**

**Correct answer:**

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

**Possible Answers:**

None of the other answers

**Correct answer:**

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

**Possible Answers:**

None of the other answers

**Correct answer:**

Rearranging the following equation

This satisfies the test of exactness, so integrating we have

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