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:

\(\displaystyle \lambda^2 - 4\lambda + 8 = 0\) We then solve the characteristic equation and find that\(\displaystyle (\lambda - 2)^2 + 4 = 0; \lambda = 2 \pm 2i\) (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  \(\displaystyle e^{2t}\sin(2t), e^{2t}\cos(2t)\).

As the given problem was homogeneous, the solution is just a linear combination of these functions. Thus, \(\displaystyle y = c_1e^{2t}\sin(2t) + c_2e^{2t}\cos(2t)\). Plugging in our initial condition, we find that \(\displaystyle 1 = c_2\). To plug in the second initial condition, we take the derivative and find that \(\displaystyle y' = 2c_1e^{2t}(\sin(2t) + \cos(2t)) + 2e^{2t}(\cos(2t) - sin(2t))\). Plugging in the second initial condition yields \(\displaystyle 2 = 2c_1 + 2\). Solving this simple system of linear equations shows us that 

\(\displaystyle c_1 = 0; c_2 = 1\)

Leaving us with a final answer of \(\displaystyle y = e^{2t}\cos(2t)\)

(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:

\(\displaystyle \lambda^3 - \lambda^2 + \lambda - 1 = 0\) 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.

\(\displaystyle \lambda^3 - \lambda^2 + \lambda - 1 = (\lambda^3 + \lambda) - (\lambda^2 + 1) = \lambda(\lambda^2 + 1) - (\lambda^2 +1) = (\lambda^2 + 1)(\lambda - 1)\)

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

\(\displaystyle 1 \left |\begin{matrix}1 && -1 && 1 && -1\\ && 1 && 0 && 1 \\ \hline \end{matrix}\right.\)

     \(\displaystyle 1 \: \, \: \: \: \, \, \: \: \: 0 \: \: \, \, \: \: \: \, \: \:1 \: \: \, \, \: \: \: \, \: \: 0\)

Which tells us the the polynomial factors into \(\displaystyle (\lambda - 1)(\lambda^2 + 1)\) and that \(\displaystyle \lambda = 1, \pm i\). This means that the fundamental set of solutions is \(\displaystyle e^t, sin(t), cos(t)\)

As the given problem was homogeneous, the solution is just a linear combination of these functions. Thus, \(\displaystyle y = c_1e^{t} + c_2\sin(t) + c_3\cos(t)\). 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:

\(\displaystyle \lambda^2 - 4\lambda - 5 = 0\) We then solve the characteristic equation and find that\(\displaystyle (\lambda - 5)(\lambda + 1) = 0 ; \lambda = 5, -1\)  This lets us know that the basis for the fundamental set of solutions to this problem (solutions to the homogeneous problem) contains  \(\displaystyle e^{5t}, e^{-t}\).

As the given problem was homogeneous, the solution is just a linear combination of these functions. Thus, \(\displaystyle y = c_1e^{5t} + c_2e^{-t}\). Plugging in our initial condition, we find that \(\displaystyle 0 = c_1 + c_2\). To plug in the second initial condition, we take the derivative and find that \(\displaystyle y' = 5c_1e^{5t} - c_2e^{-t}\). Plugging in the second initial condition yields \(\displaystyle 6 = 5c_1 - c_2\). Solving this simple system of linear equations shows us that 

\(\displaystyle c_1 = 1; c_2 = -1\)

Leaving us with a final answer of \(\displaystyle y = e^{5t} - e^{-t}\)

(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 \(\displaystyle r^2 -4r+4=0\).

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.

\(\displaystyle y= c_1e^{rt}+c_2te^{rt}\)  (for real repeated roots)

Thus, the solution is,

\(\displaystyle y(t) = c_{1}e^{2t}+c_2te^{2t}\)

This differential equation has a characteristic equation of

\(\displaystyle r^2 -5r+6=0\) , 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:

\(\displaystyle y(t)= c_1e^{r_1t}+c_2e^{r_2t}\)

with r being the roots of the characteristic equation.

Thus, the solution is

\(\displaystyle y(t)= c_1e^{2t}+c_2e^{3t}\)

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

\(\displaystyle y'' +y'-12y = 0\) .

This differential equation thus has characteristic equation of 

\(\displaystyle r^2+r-12=0\).

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

\(\displaystyle y(t)= c_1e^{3t}+c_2e^{-4t}\)

This differential equation has characteristic equation of:

\(\displaystyle r^2 - 10r + 25\) It must be noted that this characteristic equation has a double root of r=5. 

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

This is equation is

\(\displaystyle y(t)= c_1e^{r_1t}+c_2te^{r_2t}\)  in the case of a repeated root such as this, \(\displaystyle r_1 = r_2\) and is the repeated root r=5.

Therefore, the solution is

\(\displaystyle y(t)= c_1e^{5t}+c_2te^{5t}\) 

Solving the auxiliary equation

\(\displaystyle r^3+2r^2-11r-12 = 0\)

Trying out candidates for roots from the Rational Root Theorem we have a root \(\displaystyle r = -1\).

Factoring completely we have

\(\displaystyle (r+1)(r^2+r-12) = (r+4)(r-3)(r+1)=0\)

Our general solution is

\(\displaystyle y(t) = C_1e^{3x} + C_2e^{-4x} + C_3e^{-x}\)

where \(\displaystyle C_1, C_2, C_3\) are arbitrary constants.