This question is perfectly straight forward.

y"-5y'+6y=0

First find the General Solution:

\(\displaystyle \lambda ^2 -5 \lambda + 6 = 0 \,\, \Rightarrow

(\lambda -2)(\lambda-3) = 0 \\

\text{So, } \lambda_1 = 2 \text{ and } \lambda_2 = 3

\\

\\\)

Since these zeros of the characteristic equation are different, we use the general form:

\(\displaystyle \\ \\

y(x) = C_1 e^{\lambda_1 x} + C_2 e^{\lambda_2 x} \\ y(x) = C_1 e^{2 x} + C_2 e^{3 x}\)

Now we want a particular solution, so we will use your given IVs that:

1. y(0) = 1

2. y'(0) = -2

So:

\(\displaystyle y'(x) = 2C_1 e^{2x} + 3C_2e^{3x} \\

y'(0) = 2C_1 + 3C_2 = -2 \\

C_1 = \frac{1}{2} (-3C_2-2)\)

Also,

\(\displaystyle y(0) = C_1 + C_2 = 1\\

C_1 = 1-C_2\\ \\ \\\)

Therefore, \\

\(\displaystyle 1-C_2 = \frac{1}{2} (-3C_2-2)\\

2-2C_2 = -3C_2-2 \\

2+C_2 = -2 \\

C_2 = -4 \\

\\

So, \,\, C_1 = 5\)

Finally, we have:

\(\displaystyle y(x) = 5e^{2x} - 4e^{3x}\)

Note: Part C is just as easy and is also perfectly straightforward, but is non-homogeneous, so it requires a couple extra steps. Please try this one yourself and post if you have any difficulties.