115k views
13 votes
Section 5.2 Problem 13:

Solve the initial value problem.

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

y(0) = 2

y'(0) = - 17


1 Answer

6 votes

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)}

User Amuniz
by
7.5k points

Related questions

asked Feb 12, 2024 36.8k views
Ugur asked Feb 12, 2024
by Ugur
8.3k points
1 answer
3 votes
36.8k views
asked Nov 18, 2024 158k views
Ankit Mori asked Nov 18, 2024
by Ankit Mori
7.8k points
2 answers
3 votes
158k views