Question

Find the solution to initial value problem d^2y d t^2-10 d y d t+25 y=0, y(0)=6, y^(0)=2 y(t)=

Answer 1

The solution to the initial value problem is found by solving the characteristic equation of the differential equation and applying the initial conditions to determine the constants in the general solution form.

Step-by-step explanation:

The given initial value problem is a second-order linear homogeneous differential equation. To find the solution, y(t), we make use of the characteristic equation which is formed from the coefficients of the differential equation.

In our problem, the differential equation is d2y/dt2 - 10 dy/dt + 25y = 0, with initial conditions y(0) = 6, and y'(0) = 2. The characteristic equation for this is r2 - 10r + 25 = 0. Solving this quadratic equation, we get a repeated real root r = 5.

Since we have a repeated root, the general solution is of the form y(t) = (C1 + C2t)e5t. To determine the constants C1 and C2, we use the initial conditions provided, y(0) = 6 and y'(0) = 2.

• Plugging y(0) = 6 into the general solution gives us C1 = 6.
• To find C2, we differentiate the general solution to get y'(t) = C2e5t + 5(C1 + C2t)e5t and then use y'(0) = 2 to solve for C2.

Thus, the solution to the initial value problem is y(t) = (6 + C2t)e5t, where C2 is found from the initial conditions.