Variation of Parameters - Differential Equations
Card 1 of 16
Using Variation of Parameters compute the Wronskian of the following equation.
Using Variation of Parameters compute the Wronskian of the following equation.
Tap to reveal answer
To compute the Wronskian first calculates the roots of the homogeneous portion.
Therefore one of the complimentary solutions is in the form,
where,
Next compute the Wronskian:
Now take the determinant to finish calculating the Wronskian.
To compute the Wronskian first calculates the roots of the homogeneous portion.
Therefore one of the complimentary solutions is in the form,
where,
Next compute the Wronskian:
Now take the determinant to finish calculating the Wronskian.
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Using Variation of Parameters compute the Wronskian of the following equation.
Using Variation of Parameters compute the Wronskian of the following equation.
Tap to reveal answer
To compute the Wronskian first calculates the roots of the homogeneous portion.
Therefore one of the complimentary solutions is in the form,
where,
Next compute the Wronskian:
Now take the determinant to finish calculating the Wronskian.
To compute the Wronskian first calculates the roots of the homogeneous portion.
Therefore one of the complimentary solutions is in the form,
where,
Next compute the Wronskian:
Now take the determinant to finish calculating the Wronskian.
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Solve the following non-homogeneous differential equation. 
Solve the following non-homogeneous differential equation.
Tap to reveal answer
Because the inhomogeneity does not take a form we can exploit with undetermined coefficients, we must use variation of parameters. Thus, first we find the complementary solution. The characteristic equation of 
is 
, with solutions of 
. This means that 
and 
.
To do variation of parameters, we will need the Wronskian, 
Variation of parameters tells us that the coefficient in front of 
is 
where 
is the Wronskian with the 
row replaced with all 0's and a 1 at the bottom. In the 2x2 case this means that
. Plugging in, the first half simplifies to
and the second half becomes 
Putting these together with the complementary solution, we have a general solution of 
Because the inhomogeneity does not take a form we can exploit with undetermined coefficients, we must use variation of parameters. Thus, first we find the complementary solution. The characteristic equation of
To do variation of parameters, we will need the Wronskian,
Variation of parameters tells us that the coefficient in front of
and the second half becomes
Putting these together with the complementary solution, we have a general solution of
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Find a general solution to the following ODE
Find a general solution to the following ODE
Tap to reveal answer
We know the solution consists of a homogeneous solution and a particular solution.
The auxiliary equation for the homogeneous solution is
The homogeneous solution is
The particular solution is of the form
It requires variation of parameters to solve
Solving the system gets us
Integrating gets us
So 
Our solution is
We know the solution consists of a homogeneous solution and a particular solution.
The auxiliary equation for the homogeneous solution is
The homogeneous solution is
The particular solution is of the form
It requires variation of parameters to solve
Solving the system gets us
Integrating gets us
So
Our solution is
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Using Variation of Parameters compute the Wronskian of the following equation.
Using Variation of Parameters compute the Wronskian of the following equation.
Tap to reveal answer
To compute the Wronskian first calculates the roots of the homogeneous portion.
Therefore one of the complimentary solutions is in the form,
where,
Next compute the Wronskian:
Now take the determinant to finish calculating the Wronskian.
To compute the Wronskian first calculates the roots of the homogeneous portion.
Therefore one of the complimentary solutions is in the form,
where,
Next compute the Wronskian:
Now take the determinant to finish calculating the Wronskian.
← Didn't Know|Knew It →
Using Variation of Parameters compute the Wronskian of the following equation.
Using Variation of Parameters compute the Wronskian of the following equation.
Tap to reveal answer
To compute the Wronskian first calculates the roots of the homogeneous portion.
Therefore one of the complimentary solutions is in the form,
where,
Next compute the Wronskian:
Now take the determinant to finish calculating the Wronskian.
To compute the Wronskian first calculates the roots of the homogeneous portion.
Therefore one of the complimentary solutions is in the form,
where,
Next compute the Wronskian:
Now take the determinant to finish calculating the Wronskian.
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Solve the following non-homogeneous differential equation.
Solve the following non-homogeneous differential equation.
Tap to reveal answer
Because the inhomogeneity does not take a form we can exploit with undetermined coefficients, we must use variation of parameters. Thus, first we find the complementary solution. The characteristic equation of is , with solutions of . This means that and .
To do variation of parameters, we will need the Wronskian,
Variation of parameters tells us that the coefficient in front of is where is the Wronskian with the row replaced with all 0's and a 1 at the bottom. In the 2x2 case this means that
. Plugging in, the first half simplifies to
and the second half becomes
Putting these together with the complementary solution, we have a general solution of
Because the inhomogeneity does not take a form we can exploit with undetermined coefficients, we must use variation of parameters. Thus, first we find the complementary solution. The characteristic equation of
To do variation of parameters, we will need the Wronskian,
Variation of parameters tells us that the coefficient in front of
and the second half becomes
Putting these together with the complementary solution, we have a general solution of
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Find a general solution to the following ODE
Find a general solution to the following ODE
Tap to reveal answer
We know the solution consists of a homogeneous solution and a particular solution.
The auxiliary equation for the homogeneous solution is
The homogeneous solution is
The particular solution is of the form
It requires variation of parameters to solve
Solving the system gets us
Integrating gets us
So
Our solution is
We know the solution consists of a homogeneous solution and a particular solution.
The auxiliary equation for the homogeneous solution is
The homogeneous solution is
The particular solution is of the form
It requires variation of parameters to solve
Solving the system gets us
Integrating gets us
So
Our solution is
← Didn't Know|Knew It →
Using Variation of Parameters compute the Wronskian of the following equation.
Using Variation of Parameters compute the Wronskian of the following equation.
Tap to reveal answer
To compute the Wronskian first calculates the roots of the homogeneous portion.
Therefore one of the complimentary solutions is in the form,
where,
Next compute the Wronskian:
Now take the determinant to finish calculating the Wronskian.
To compute the Wronskian first calculates the roots of the homogeneous portion.
Therefore one of the complimentary solutions is in the form,
where,
Next compute the Wronskian:
Now take the determinant to finish calculating the Wronskian.
← Didn't Know|Knew It →
Using Variation of Parameters compute the Wronskian of the following equation.
Using Variation of Parameters compute the Wronskian of the following equation.
Tap to reveal answer
To compute the Wronskian first calculates the roots of the homogeneous portion.
Therefore one of the complimentary solutions is in the form,
where,
Next compute the Wronskian:
Now take the determinant to finish calculating the Wronskian.
To compute the Wronskian first calculates the roots of the homogeneous portion.
Therefore one of the complimentary solutions is in the form,
where,
Next compute the Wronskian:
Now take the determinant to finish calculating the Wronskian.
← Didn't Know|Knew It →
Solve the following non-homogeneous differential equation.
Solve the following non-homogeneous differential equation.
Tap to reveal answer
Because the inhomogeneity does not take a form we can exploit with undetermined coefficients, we must use variation of parameters. Thus, first we find the complementary solution. The characteristic equation of is , with solutions of . This means that and .
To do variation of parameters, we will need the Wronskian,
Variation of parameters tells us that the coefficient in front of is where is the Wronskian with the row replaced with all 0's and a 1 at the bottom. In the 2x2 case this means that
. Plugging in, the first half simplifies to
and the second half becomes
Putting these together with the complementary solution, we have a general solution of
Because the inhomogeneity does not take a form we can exploit with undetermined coefficients, we must use variation of parameters. Thus, first we find the complementary solution. The characteristic equation of
To do variation of parameters, we will need the Wronskian,
Variation of parameters tells us that the coefficient in front of
and the second half becomes
Putting these together with the complementary solution, we have a general solution of
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Find a general solution to the following ODE
Find a general solution to the following ODE
Tap to reveal answer
We know the solution consists of a homogeneous solution and a particular solution.
The auxiliary equation for the homogeneous solution is
The homogeneous solution is
The particular solution is of the form
It requires variation of parameters to solve
Solving the system gets us
Integrating gets us
So
Our solution is
We know the solution consists of a homogeneous solution and a particular solution.
The auxiliary equation for the homogeneous solution is
The homogeneous solution is
The particular solution is of the form
It requires variation of parameters to solve
Solving the system gets us
Integrating gets us
So
Our solution is
← Didn't Know|Knew It →
Using Variation of Parameters compute the Wronskian of the following equation.
Using Variation of Parameters compute the Wronskian of the following equation.
Tap to reveal answer
To compute the Wronskian first calculates the roots of the homogeneous portion.
Therefore one of the complimentary solutions is in the form,
where,
Next compute the Wronskian:
Now take the determinant to finish calculating the Wronskian.
To compute the Wronskian first calculates the roots of the homogeneous portion.
Therefore one of the complimentary solutions is in the form,
where,
Next compute the Wronskian:
Now take the determinant to finish calculating the Wronskian.
← Didn't Know|Knew It →
Using Variation of Parameters compute the Wronskian of the following equation.
Using Variation of Parameters compute the Wronskian of the following equation.
Tap to reveal answer
To compute the Wronskian first calculates the roots of the homogeneous portion.
Therefore one of the complimentary solutions is in the form,
where,
Next compute the Wronskian:
Now take the determinant to finish calculating the Wronskian.
To compute the Wronskian first calculates the roots of the homogeneous portion.
Therefore one of the complimentary solutions is in the form,
where,
Next compute the Wronskian:
Now take the determinant to finish calculating the Wronskian.
← Didn't Know|Knew It →
Solve the following non-homogeneous differential equation.
Solve the following non-homogeneous differential equation.
Tap to reveal answer
Because the inhomogeneity does not take a form we can exploit with undetermined coefficients, we must use variation of parameters. Thus, first we find the complementary solution. The characteristic equation of is , with solutions of . This means that and .
To do variation of parameters, we will need the Wronskian,
Variation of parameters tells us that the coefficient in front of is where is the Wronskian with the row replaced with all 0's and a 1 at the bottom. In the 2x2 case this means that
. Plugging in, the first half simplifies to
and the second half becomes
Putting these together with the complementary solution, we have a general solution of
Because the inhomogeneity does not take a form we can exploit with undetermined coefficients, we must use variation of parameters. Thus, first we find the complementary solution. The characteristic equation of
To do variation of parameters, we will need the Wronskian,
Variation of parameters tells us that the coefficient in front of
and the second half becomes
Putting these together with the complementary solution, we have a general solution of
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Find a general solution to the following ODE
Find a general solution to the following ODE
Tap to reveal answer
We know the solution consists of a homogeneous solution and a particular solution.
The auxiliary equation for the homogeneous solution is
The homogeneous solution is
The particular solution is of the form
It requires variation of parameters to solve
Solving the system gets us
Integrating gets us
So
Our solution is
We know the solution consists of a homogeneous solution and a particular solution.
The auxiliary equation for the homogeneous solution is
The homogeneous solution is
The particular solution is of the form
It requires variation of parameters to solve
Solving the system gets us
Integrating gets us
So
Our solution is
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