221k views
4 votes
Find the solution of the initial value problem: y" - y' - 30y = 0, y(0) = 2.

User Darmat
by
7.7k points

1 Answer

4 votes

Final answer:

To solve the given differential equation y" - y' - 30y = 0 with the initial condition y(0) = 2, we assume a solution of the form y = e^(rt). Substituting this into the differential equation, factoring out e^(rt), and solving the resulting quadratic equation, we find the general solution y = c1e^(-5t) + c2e^(6t). Using the initial condition y(0) = 2, we can solve for the constants c1 and c2 to get the specific solution.

Step-by-step explanation:

The given initial value problem is y" - y' - 30y = 0, y(0) = 2.

To solve this differential equation, we can assume a solution of the form y = e^(rt).

Substituting this into the differential equation, we get r^2 e^(rt) - re^(rt) - 30e^(rt) = 0.

Factoring out e^(rt), we get e^(rt)(r^2 - r - 30) = 0.

Solving the quadratic equation r^2 - r - 30 = 0, we get r = -5 or r = 6.

Therefore, the general solution to the differential equation is y = c1e^(-5t) + c2e^(6t), where c1 and c2 are constants determined by the initial conditions.

Using the initial condition y(0) = 2, we can substitute t = 0 and y = 2 into the general solution to solve for the constants c1 and c2.

When t = 0, y = c1e^(0) + c2e^(0) = c1 + c2 = 2.

Since the equation c1 + c2 = 2 has infinitely many solutions, we need an additional condition to uniquely determine the values of c1 and c2.

User James Iry
by
8.2k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories