Putting the equation in standard form we get that

$\displaystyle y^{'''} - \frac{3}{(x-1)}y^{''} + \frac{6x}{(x-1)}y' - \frac{\cos{x}}{x(x-1)}y = \frac{\sqrt{x+5}}{x(x-1)}$

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

$\displaystyle (-5,0), (0,1), (1,\infty)$. The choice that is not a subset of these is $\displaystyle (0,2)$

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. 

First solve the homogeneous portion:

$\displaystyle \\y''-8y'+16y=0 \\m^2-8m+16=0 \\(m-4)(m-4)=0$

Therefore, $\displaystyle m=4$ is a repeated root thus one of the complimentary solutions $\displaystyle y_c$ is,

$\displaystyle y_c=c_1e^{4x}+c_2xe^{4x}$

Now find the remaining complimentary solution $\displaystyle y_p$.

$\displaystyle \\y_p=Ax+B \\y_p'=A \\y_p''=0$

Now solve for $\displaystyle A$ and $\displaystyle B$.

$\displaystyle \\0-8A+16(Ax+B)=24x+2 \\-8A+16Ax+16B=24x+2$

Where 

$\displaystyle \\16A=24 \\\\A=\frac{24}{16}=\frac{3}{2}$

and 

$\displaystyle \\-8A+16B=2 \\\\-8\left(\frac{3}{2} \right )+16B=2 \\\\-\frac{24}{2}+16B=2 \\\\-12+16B=2 \\\\16B=14 \\\\B=\frac{14}{16}=\frac{7}{8}$

Therefore,

$\displaystyle y_p=\frac{3}{2}x+\frac{7}{8}$

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

$\displaystyle \\y=y_c+y_p \\y=c_1e^{4x}+c_2xe^{4x}+\frac{3}{2}x+\frac{7}{8}$