Find the unique solution to the differential equation

y"?1y'?6y=?24(t^2)+10t+41
satisfying the initial conditions y(0)=?6 and y'(0)=?26

To find the unique solution to the given differential equation with the given initial conditions, we will use the method of solving linear, homogeneous second-order differential equations.

Step 1: Write the characteristic equation
The characteristic equation for the given differential equation is:
r^2 - r - 6 = 0

Step 2: Solve the characteristic equation
To find the values of r, solve the characteristic equation. Factoring the equation, we have:
(r - 3)(r + 2) = 0

So, the solutions are r = 3 and r = -2.

Step 3: Write the general solution of the homogeneous equation
The general solution of the homogeneous equation is:
y_h(t) = c1*e^(3t) + c2*e^(-2t)

Step 4: Find the particular solution of the non-homogeneous equation
Since the right side of the equation is non-homogeneous, we need to find a particular solution. We'll assume a particular solution of the form:
y_p(t) = At^2 + Bt + C

Taking its derivatives, we have:
y_p'(t) = 2At + B
y_p''(t) = 2A

Substituting these derivatives and the particular solution back into the differential equation, we get:
2A - (2At + B) - 6(At^2 + Bt + C) = 24t^2 + 10t + 41

Simplifying and equating coefficients, we have:
-6A = 24 (coefficient of t^2)
-2A - 6B = 10 (coefficient of t)
2A - B - 6C = 41 (constant term)

Solving these equations, we find A = -4, B = -7, and C = -3.

Therefore, the particular solution is:
y_p(t) = -4t^2 - 7t - 3

Step 5: Write the general solution of the non-homogeneous equation
The general solution of the non-homogeneous equation is the sum of the general solution of the homogeneous equation and the particular solution:
y(t) = y_h(t) + y_p(t)
= c1*e^(3t) + c2*e^(-2t) - 4t^2 - 7t - 3

Step 6: Apply initial conditions to find the unique solution
Using the given initial conditions, we can determine the values of c1 and c2.

Given: y(0) = -6 and y'(0) = -26

Substituting t = 0 and y(0) = -6 into the general solution:
-6 = c1 + c2 - 3

Substituting t = 0 and y'(0) = -26 into the derivative of the general solution:
-26 = 3c1 - 2c2 - 7

Solving these two equations simultaneously, we find c1 = -3 and c2 = -10.

Therefore, the unique solution to the given differential equation with the given initial conditions is:
y(t) = -3e^(3t) - 10e^(-2t) - 4t^2 - 7t - 3