Linear & Exact Equations - Differential Equations
Card 0 of 36
Find the general solution of the given differential equation and determine if there are any transient terms in the general solution.

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.
![\\int \frac{d}{dx}[x^3y]=\int x^3(x^3-1)dx \\x^3y=\int x^6-x^3dx \\x^3y=\frac{x^7}{7}-\frac{x^4}{4}+c](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/1002873/gif.latex)
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
.
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 .
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Find the solution for the following differential equation:
where
.
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
.
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 .
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Consider the differential equation

Which of the terms in the differential equation make the equation nonlinear?
Consider the differential equation
Which of the terms in the differential equation make the equation nonlinear?
The term
makes the differential equation nonlinear because a linear equation has the form of

The term makes the differential equation nonlinear because a linear equation has the form of
Compare your answer with the correct one above
Find the general solution of the given differential equation and determine if there are any transient terms in the general solution.

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.
![\\int \frac{d}{dx}[x^3y]=\int x^3(x^3-1)dx \\x^3y=\int x^6-x^3dx \\x^3y=\frac{x^7}{7}-\frac{x^4}{4}+c](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/1002873/gif.latex)
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 
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
Compare your answer with the correct one above
Is the following differential equation exact? 
If so, find the general solution.
Is the following differential equation exact?
If so, find the general solution.
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.
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.
Compare your answer with the correct one above
Is the following differential equation exact?
If so, find the general solution.
Is the following differential equation exact? If so, find the general solution.
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.
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.
Compare your answer with the correct one above
Solve the Following Equation

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

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
Compare your answer with the correct one above
Find the general solution of the differential equation

Find the general solution of the differential equation
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 

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
Compare your answer with the correct one above
Solve the differential equation

Solve the differential equation
Rearranging the following equation

This satisfies the test of exactness, so integrating we have

Rearranging the following equation
This satisfies the test of exactness, so integrating we have
Compare your answer with the correct one above
Find the general solution of the given differential equation and determine if there are any transient terms in the general solution.

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.
![\\int \frac{d}{dx}[x^3y]=\int x^3(x^3-1)dx \\x^3y=\int x^6-x^3dx \\x^3y=\frac{x^7}{7}-\frac{x^4}{4}+c](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/1002873/gif.latex)
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
.
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 .
Compare your answer with the correct one above
Find the solution for the following differential equation:
where
.
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
.
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 .
Compare your answer with the correct one above
Consider the differential equation

Which of the terms in the differential equation make the equation nonlinear?
Consider the differential equation
Which of the terms in the differential equation make the equation nonlinear?
The term
makes the differential equation nonlinear because a linear equation has the form of

The term makes the differential equation nonlinear because a linear equation has the form of
Compare your answer with the correct one above
Find the general solution of the given differential equation and determine if there are any transient terms in the general solution.

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.
![\\int \frac{d}{dx}[x^3y]=\int x^3(x^3-1)dx \\x^3y=\int x^6-x^3dx \\x^3y=\frac{x^7}{7}-\frac{x^4}{4}+c](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/1002873/gif.latex)
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 
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
Compare your answer with the correct one above
Is the following differential equation exact? 
If so, find the general solution.
Is the following differential equation exact?
If so, find the general solution.
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.
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.
Compare your answer with the correct one above
Is the following differential equation exact?
If so, find the general solution.
Is the following differential equation exact? If so, find the general solution.
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.
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.
Compare your answer with the correct one above
Solve the Following Equation

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

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
Compare your answer with the correct one above
Find the general solution of the differential equation

Find the general solution of the differential equation
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 

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
Compare your answer with the correct one above
Solve the differential equation

Solve the differential equation
Rearranging the following equation

This satisfies the test of exactness, so integrating we have

Rearranging the following equation
This satisfies the test of exactness, so integrating we have
Compare your answer with the correct one above
Find the general solution of the given differential equation and determine if there are any transient terms in the general solution.

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.
![\\int \frac{d}{dx}[x^3y]=\int x^3(x^3-1)dx \\x^3y=\int x^6-x^3dx \\x^3y=\frac{x^7}{7}-\frac{x^4}{4}+c](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/1002873/gif.latex)
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
.
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 .
Compare your answer with the correct one above
Find the solution for the following differential equation:
where
.
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
.
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 .
Compare your answer with the correct one above