Find the general solution of (D^2-3D+2)(D-5)^Dy=0. (Hint:Use the auxiliary equation)

To find the general solution of the given differential equation, we first need to solve the auxiliary equation.

The auxiliary equation is obtained by setting the given equation equal to zero, i.e., (D^2-3D+2)(D-5)^Dy=0.

Step 1: Solve the auxiliary equation (D^2-3D+2)(D-5)^Dy=0.

To solve the auxiliary equation, we consider each factor separately:

For the factor D^2-3D+2=0, we can use the quadratic formula or factoring to solve for D. In this case, we can factor the expression as (D-2)(D-1)=0. Therefore, we get two possible values for D: D = 2 and D = 1.

For the factor (D-5)^Dy=0, we can see that it represents a repeated root. The repeated root is D = 5.

Step 2: Write down the general solution.

The general solution of a linear homogeneous differential equation can be written in the form:

y = C1e^(r1x) + C2xe^(r1x) + C3e^(r2x) + C4xe^(r2x) + ... + Cke^(rx) + Ck+1xe^(rx) + ...

where r1, r2, ..., rk are the distinct roots of the auxiliary equation, and C1, C2, ..., Ck are constants.

In our case, the auxiliary equation has two distinct roots (D = 2 and D = 1) and one repeated root (D = 5). Therefore, the general solution can be expressed as:

y = C1e^(2x) + C2xe^(2x) + C3e^(x) + C4xe^(x) + C5e^(5x) + C6xe^(5x) + ...

where C1, C2, C3, C4, C5, C6, ... are constants.

Note: The general solution accounts for all possible solutions of the differential equation. The constants C1, C2, C3, C4, C5, C6, ... can be determined using initial/boundary conditions (if given) or through further analysis of the problem.