Section 5.2 Problem 13:

Solve the initial value problem.
[tex]y'' + 14y' + 50y = 0[/tex]
[tex]y(0) = 2[/tex]
[tex]y'(0) = - 17[/tex]

Respuesta :

The characteristic equation is

[tex]r^2 + 14r + 50 = 0[/tex]

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

[tex]y_c = C_1 \cos(7x) + C_2 \sin(7x)[/tex]

whose derivative is

[tex]{y_c}' = -7C_1 \sin(7x) + 7C_2 \cos(7x)[/tex]

Use the initial conditions to solve for the constants:

[tex]y(0) = 2 \implies 2 = C_1[/tex]

[tex]y'(0) = -17 \implies -17 = 7C_2 \implies C_2 = -\dfrac{17}7[/tex]

Then the particular solution is

[tex]\boxed{y(x) = 2 \cos(7x) - \dfrac{17}7 \sin(7x)}[/tex]