### All Differential Equations Resources

## Example Questions

### Example Question #1 : Variation Of Parameters

Using Variation of Parameters compute the Wronskian of the following equation.

**Possible Answers:**

**Correct 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.

### Example Question #1 : Variation Of Parameters

Using Variation of Parameters compute the Wronskian of the following equation.

**Possible Answers:**

**Correct 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.

### Example Question #1 : Variation Of Parameters

Solve the following non-homogeneous differential equation.

**Possible Answers:**

**Correct 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

### Example Question #2 : Variation Of Parameters

Find a general solution to the following ODE

**Possible Answers:**

None of the other solutions

**Correct 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

Certified Tutor

Certified Tutor