Section 5.2 Problem 13: Solve the initial value problem. y'' + 14y' + 50y = 0 y(0) = 2 y'(0) = - 17 ​

Question

Section 5.2 Problem 13:

Solve the initial value problem.

`y'' + 14y' + 50y = 0`

`y(0) = 2`

`y'(0) = - 17`
​

Answer 1

The characteristic equation is

`r^2 + 14r + 50 = 0`

with complex roots r = -7 ± i, so the characteristic solution is

`y_c = C_1 \cos(7x) + C_2 \sin(7x)`

whose derivative is

`{y_c}' = -7C_1 \sin(7x) + 7C_2 \cos(7x)`

Use the initial conditions to solve for the constants:

`y(0) = 2 \implies 2 = C_1`

`y'(0) = -17 \implies -17 = 7C_2 \implies C_2 = -\frac{17}7`

Then the particular solution is

`\boxed{y(x) = 2 \cos(7x) - \frac{17}7 \sin(7x)}`