(1 pt) Find the unique solution to the differential equation

y"?5y'+6y=?312sin(2t)
satisfying the initial conditions y(0)=?31 and y'(0)=?18

To solve the differential equation y'' - 5y' + 6y = 312sin(2t) with the given initial conditions y(0) = -31 and y'(0) = -18, we will use the method of undetermined coefficients.

First, find the complementary solution by solving the homogeneous equation y'' - 5y' + 6y = 0.

The characteristic equation is r^2 - 5r + 6 = 0.

Factoring, we have (r - 2)(r - 3) = 0.

So the complementary solution is y_c(t) = c1e^(2t) + c2e^(3t), where c1 and c2 are arbitrary constants.

Next, find the particular solution using the method of undetermined coefficients. Since the right-hand side of the equation is sin(2t), we will assume a particular solution of the form y_p(t) = A sin(2t) + B cos(2t), where A and B are coefficients to be determined.

Taking the derivatives, we have y_p'(t) = 2A cos(2t) - 2B sin(2t) and y_p''(t) = -4A sin(2t) - 4B cos(2t).

Substituting these derivatives into the differential equation, we have:

(-4A sin(2t) - 4B cos(2t)) - 5(2A cos(2t) - 2B sin(2t)) + 6(A sin(2t) + B cos(2t)) = 312sin(2t).

Simplifying, we get:

(-4A - 10B + 6A)sin(2t) + (-4B + 10A + 6B)cos(2t) = 312sin(2t).

Equating the coefficients of sin(2t) and cos(2t) on both sides, we get two equations:

-4A - 10B + 6A = 312 and -4B + 10A + 6B = 0.

Solving these equations, we find A = -39 and B = 24.

Therefore, the particular solution is y_p(t) = -39sin(2t) + 24cos(2t).

Finally, the general solution is y(t) = y_c(t) + y_p(t) = c1e^(2t) + c2e^(3t) - 39sin(2t) + 24cos(2t).

To find the unique solution satisfying the initial conditions, substitute the values of t and the initial conditions into the general solution.

y(0) = c1 + c2 - 39 = -31,

y'(0) = 2c1 + 3c2 + 24 = -18.

Solving these equations simultaneously, we find c1 = -15 and c2 = 23.

Therefore, the unique solution to the differential equation satisfying the initial conditions is y(t) = -15e^(2t) + 23e^(3t) - 39sin(2t) + 24cos(2t).

To find the unique solution to the given differential equation, we can follow these steps:

Step 1: Find the characteristic equation.
The characteristic equation is obtained by setting the coefficients of the differential equation to zero. In this case, the characteristic equation is:
r^2 - 5r + 6 = 0

Step 2: Solve the characteristic equation.
We can solve the characteristic equation by factoring or using the quadratic formula. Factoring the equation, we have:
(r - 2)(r - 3) = 0

This yields two solutions: r = 2 and r = 3.

Step 3: Write the homogeneous solution.
The homogeneous solution is obtained by multiplying the solutions from Step 2 with arbitrary constants:
y_h(t) = C1e^(2t) + C2e^(3t)

Step 4: Find the particular solution.
In order to find the particular solution, we need to consider the right-hand side of the differential equation. In this case, it is 312sin(2t).

Since sin(2t) has the same form as the solution, we need a particular solution of the form At*cos(2t) + Bt*sin(2t). Differentiating this particular solution twice and substituting it into the differential equation, we can solve for A and B.

Step 5: Apply the initial conditions.
Using the given initial conditions y(0) = -31 and y'(0) = -18, we substitute these values into the general solution:
y(0) = C1e^(2*0) + C2e^(3*0) + A(0)*cos(2*0) + B(0)*sin(2*0) = -31
y'(0) = 2C1e^(2*0) + 3C2e^(3*0) - 2A(0)*sin(2*0) + B(0)*cos(2*0) = -18

Simplifying these equations, we obtain two equations:
C1 + C2 = -31
2C1 + 3C2 + B = -18

Step 6: Solve for the arbitrary constants.
Solve the equations obtained in Step 5 to find the values of C1, C2, A, and B.

Step 7: Write the unique solution.
Now that we have the values of the arbitrary constants, substitute them into the general solution to obtain the unique solution.

Therefore, by following these steps, we can find the unique solution to the given differential equation satisfying the initial conditions.