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.
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
.
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.
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.
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.
Yes.
No. The equation does not take the proper form.
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.
Yes.
Yes.
No. The equation does not take the right form.
Yes.
No. The partial derivatives in the equation do not have the right relationship.
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
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
None of the other answers
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
None of the other answers
Rearranging the following equation
This satisfies the test of exactness, so integrating we have
All Differential Equations Resources
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