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．插入第二个初始条件,我们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 intoand 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.

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 thatThis 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．插入第二个初始条件,我们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)

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 thehomogeneousequation 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:

This differential equation has characteristic equation of:

It must be noted that this characteristic equation has adouble 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

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

whereare arbitrary constants.