Higher-Order Differential Equations - Differential Equations

This is a linear higher order differential equation. First, we need the characteristic equation, which is just obtained by turning the derivative orders into powers to get the following:

We then solve the characteristic equation and find that This lets us know that the basis for the fundamental set of solutions to this problem (solutions to the homogeneous problem) contains .

As the given problem was homogeneous, the solution is just a linear combination of these functions. Thus, . Plugging in our initial condition, we find that . To plug in the second initial condition, we take the derivative and find that . Plugging in the second initial condition yields . Solving this simple system of linear equations shows us that

Leaving us with a final answer of

(Note, it would have been very simple to find the right answer just by taking derivatives and plugging in, but this is not overly helpful for non-multiple choice questions)

This differential equation has characteristic equation of:

It must be noted that this characteristic equation has a double rootof r=5.

Thus the general solution to a homogeneous differential equation with a repeated root is used.

This is equation is

in the case of a repeated root such as this, and is the repeated root r=5.

Therefore, the solution is

This is a linear higher order differential equation. First, we need the characteristic equation, which is just obtained by turning the derivative orders into powers to get the following:

We then solve the characteristic equation and find that (Use the quadratic formula if you'd like) This lets us know that the basis for the fundamental set of solutions to this problem (solutions to the homogeneous problem) contains .

As the given problem was homogeneous, the solution is just a linear combination of these functions. Thus, . Plugging in our initial condition, we find that . To plug in the second initial condition, we take the derivative and find that . Plugging in the second initial condition yields . Solving this simple system of linear equations shows us that

Leaving us with a final answer of

(Note, it would have been very simple to find the right answer just by taking derivatives and plugging in, but this is not overly helpful for non-multiple choice questions)

This is a linear higher order differential equation. First, we need the characteristic equation, which is just obtained by turning the derivative orders into powers to get the following:

To factor this, in this case we may use factoring by grouping. More generally, we may use horner's scheme/synthetic division to test possible roots. Here are both methods shown.

Alternatively, the rational root theorem suggests that we try -1 or 1 as a root of this equation. Using horner's scheme, we see

Which tells us the the polynomial factors into and that . This means that the fundamental set of solutions is

As the given problem was homogeneous, the solution is just a linear combination of these functions. Thus, . As this is not an initial value problem and just asks for the general solution, we are done.

The ode has a characteristic equation of .

This yields the double root of r=2. Then the roots are plugged into the general solution to a homogeneous differential equation with a repeated root.

(for real repeated roots)

Thus, the solution is,

This differential equation has a characteristic equation of

, which yields the roots for r=2 and r=3. Once the roots or established to be real and non-repeated, the general solution for homogeneous linear ODEs is used. this equation is given as:

with r being the roots of the characteristic equation.

Thus, the solution is

We start off by noting that the homogeneous equation we are trying to solve is given as

.

This differential equation thus has characteristic equation of

.

This has roots of r=3 and r=-4, therefore, the general homogeneous solution is given by:

Solving the auxiliary equation

Trying out candidates for roots from the Rational Root Theorem we have a root .

Factoring completely we have

Our general solution is

where are arbitrary constants.

Putting the equation in standard form we get that

We need to find where these coefficients are simultaneously continuous. This is where

. The choice that is not a subset of these is

First solve the homogeneous portion:

Therefore, is a repeated root thus one of the complimentary solutions is,

Now find the remaining complimentary solution .

Now solve for and .

Where

and

Therefore,

Now, combine both of the complimentary solutions together to arrive at the general solution.

This is a higher order inhomogeneous linear differential equation. Because the inhomogeneity is a cosine, we will use variation of parameters to solve it.

First, we find the characteristic equation to solve for the homogenous solution. This gives us .

This tells us that the homogeneous solution is . As neither of these overlap with our inhomogeneity, we are safe to continue without adding a factor of t.

Thus, let us guess that . Then,

and

Plugging into the original equation, we have

Which implies that and . Solving via substitution,

Thus, the particular solution is and the overall solution is the particular plus the homogeneous.

So

We first note that the differential equation has characteristic equation

,

since the roots of this characteristic polynomial are linearly independent of the forcing function

,

we simply use undetermined coefficient combination rules to figure that the particular solution will be of form

We first figure that the forcing function is linearly independent to the homogeneous solution solved with the characteristic equation.

Therefore, using proper undetermined coefficients function rules, the particular solution will be of the form:

It is important to note that when either a sine or a cosine is used, both sine and cosine must show up in the particular solution guess.

We start with the assumption that the particular solution must be of the form

.

Then we solve the first and second derivatives with this assumption, that is,

and .

Then we plug in these quantities into the given equation to get:

, which solves for .

Thus, but the method of undetermined coefficients, a particular solution to this differential equation is:

First solve the homogeneous portion:

Therefore, is a repeated root thus one of the complimentary solutions is,

Now find the remaining complimentary solution .

Now solve for and .

Where

and

Therefore,

Now, combine both of the complimentary solutions together to arrive at the general solution.

The form of a guess for a particular solution is

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.

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

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