y''-3y'+2y=0;y(0)=1;y'(0)=2

To solve the given differential equation, we can assume a solution of the form y=e^(rt), where r is a constant to be determined.

Plugging this assumed solution into the differential equation, we get:

r^2e^(rt) - 3re^(rt) + 2e^(rt) = 0

Dividing through by e^(rt), we have:

r^2 - 3r + 2 = 0

This is a quadratic equation, which can be factored as:

(r-1)(r-2) = 0

So, we have two possible values for r: r=1 and r=2.

Therefore, the general solution to the differential equation is:

y = c1e^(rt) + c2e^(rt)

Substituting the given initial conditions, we have:

y(0) = c1e^(0) + c2e^(0) = c1 + c2 = 1

y'(0) = c1re^(0) + c2re^(0) = c1r + c2r = 2

From the first equation, we can solve for c1 in terms of c2:

c1 = 1 - c2

Substituting this into the second equation, we have:

(1 - c2)r + c2r = 2

r - c2r + c2r = 2

r = 2

So, we have r=2 and c1=1-c2. Therefore, the particular solution to the differential equation is:

y = (1-c2)e^(2t) + c2e^(2t)

Using the initial condition y(0) = 1, we can solve for c2:

(1-c2)e^(0) + c2e^(0) = 1

1 - c2 + c2 = 1

1 = 1

Since this equation is true for any value of c2, we can choose any value for c2.

Using the initial condition y'(0) = 2, we can solve for c1:

(1-c2)(2e^(0)) + c2(2e^(0)) = 2

2 - 2c2 + 2c2 = 2

2 = 2

Since this equation is also true for any value of c2, we can choose any value for c2.

Therefore, the particular solution to the differential equation with the given initial conditions is:

y = (1-c2)e^(2t) + c2e^(2t)

where c2 can be any real number.