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 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 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:

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.

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,

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

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