All Differential Equations Resources
Example Questions
Example Question #1 : Homogeneous Linear Systems
Find the general solution to the given system.
To find the general solution to the given system
first find the eigenvalues and eigenvectors.
Therefore the eigenvalues are
Now calculate the eigenvectors
For
Thus,
For
Thus
Therefore,
Now the general solution is,
Example Question #1 : System Of Linear First Order Differential Equations
Solve the initial value problem . Where
To solve the homogeneous system, we will need a fundamental matrix. Specifically, it will help to get the matrix exponential. To do this, we will diagonalize the matrix. First, we will find the eigenvalues which we can do by calculating the determinant of .
Finding the eigenspaces, for lambda = 1, we have
Adding -1/2 Row 1 to Row 2 and dividing by -1/2, we have which means
Thus, we have an eigenvector of .
For lambda = 4
Adding Row 1 to Row 2, we have
So with an eigenvector
.
Thus, we have and
. Using the inverse formula for 2x2 matrices, we have that
. As we know that
, we have
The solution to a homogenous system of linear equations is simply to multiply the matrix exponential by the intial condition. For other fundamental matrices, the matrix inverse is needed as well.
Thus, our final answer is
Example Question #1 : System Of Linear First Order Differential Equations
Solve the homogenous equation:
With the initial conditions:
none of these answers
So this is a homogenous, second order differential equation. In order to solve this we need to solve for the roots of the equation. This equation can be written as:
Which, using the quadratic formula or factoring gives us roots of
and
The solution of homogenous equations is written in the form:
so we don't know the constants, but can substitute the values we solved for the roots:
We have two initial values, one for y(t) and one for y'(t), both with t=0\
So:
so:
We can solve for :
Then plug into the other equation to solve for
So, solving, we get: Then
This gives a final answer of:
Example Question #1 : Homogeneous Linear Systems
Solve the second order differential equation:
Subject to the initial values:
none of these answers
So this is a homogenous, second order differential equation. In order to solve this we need to solve for the roots of the equation. This equation can be written as:
Which, using the quadratic formula or factoring gives us roots of
and
The solution of homogenous equations is written in the form:
so we don't know the constants, but can substitute the values we solved for the roots:
We have two initial values, one for y(t) and one for y'(t), both with t=0
So:
so:
We can solve Then plug into the other equation to solve for
So, solving, we get: Then
This gives a final answer of:
Example Question #1 : Homogeneous Linear Systems
Solve the differential equation for y:
Subject to the initial condition:
So this is a homogenous, first order differential equation. In order to solve this we need to solve for the roots of the equation. This equation can be written as:
gives us a root of
The solution of homogenous equations is written in the form:
so we don't know the constant, but can substitute the values we solved for the root:
We have one initial values, for y(t) with t=0
So:
This gives a final answer of:
Example Question #1 : System Of Linear First Order Differential Equations
Solve the third order differential equation:
none of these answers
So this is a homogenous, third order differential equation. In order to solve this we need to solve for the roots of the equation. This equation can be written as:
Which, using the cubic formula or factoring gives us roots of
,
and
The solution of homogenous equations is written in the form:
so we don't know the constants, but can substitute the values we solved for the roots:
We have three initial values, one for y(t), one for y'(t), and for y''(t) all with t=0
So:
so:
So this can be solved either by substitution or by setting up a 3X3 matrix and reducing. Once you do either of these methods, the values for the constants will be: Then
and
This gives a final answer of:
Example Question #1 : System Of Linear First Order Differential Equations
Solve the differential equation:
Subject to the initial conditions:
So this is a homogenous, third order differential equation. In order to solve this we need to solve for the roots of the equation. This equation can be written as:
Which, using the cubic formula or factoring gives us roots of
,
and
The solution of homogenous equations is written in the form:
so we don't know the constants, but can substitute the values we solved for the roots:
We have three initial values, one for y(t), one for y'(t), and for y''(t) all with t=0
So:
So this can be solved either by substitution or by setting up a 3X3 matrix and reducing. Once you do either of these methods, the values for the constants will be: Then
and
This gives a final answer of:
Example Question #8 : Homogeneous Linear Systems
Find the general solution to the given system.
To find the general solution to the given system
first find the eigenvalues and eigenvectors.
Therefore the eigenvalues are
Now calculate the eigenvectors
For
Thus,
For
Thus
Therefore,
Now the general solution is,
Example Question #71 : Differential Equations
When substituted into the homogeneous linear system for
, which of the following matrices will have a saddle point equilibrium in its phase plane?
A saddle point phase plane results from two real eigenvalues of different signs. Three of these matrices are triangular, which means their eigenvalues are on the diagonal. For these three, the eigenvalues are real, but both the same sign, meaning they don't have saddles. For the remaining two, we'll need to find the eigenvalues using the characteristic equations.
For , we have
The discriminant to this is , so the solutions are non-real. Thus, this matrix doesn't yield a saddle point.
For we have,
We see that this matrix yields two real eigenvalues with different signs. Thus, it is the correct choice.
Example Question #1 : System Of Linear First Order Differential Equations
Find the general solution to the system of ordinary differential equations
where
None of the other answers.
Finding the eigenvalues and eigenvectors of with the characteristic equation of the matrix
The corresponding eigenvalues are, respectively
and
This gives us that the general solution is
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