Find-the-particular-solution-to-the-differential-equation-2y-5y-2y-0-subject-to-the-initial-conditions-y-0-2y-y-0-1-

Question Number 161462 by ZiYangLee last updated on 18/Dec/21

Find the particular solution to the differential

equation 2 y^{″} + 5 y^{'} + 2 y = 0 subject to the initial

conditions y (0) = 2 y, y^{'} (0) = 1 .

Answered by TheSupreme last updated on 18/Dec/21

2 λ^{2} + 5 λ + 2 = 0

λ_{1, 2} = \frac{- 5 \pm \sqrt{25 - 16}}{4} = \frac{- 5 \pm 3}{4} = - 2, - \frac{1}{2}

y_{0} = c_{1} e^{- 2 t} + c_{2} e^{- \frac{t}{2}}

y (0) = c_{1} + c_{2} = 2

y^{'} (0) = - 2 c_{1} - \frac{1}{2} c_{2} = 1

- 2 c_{2} + 4 - \frac{1}{2} c_{2} = 1

- \frac{5}{2} c_{2} = - 3

c_{2} = \frac{6}{5} \to c_{1} = \frac{4}{5}

y (t) = \frac{4}{5} e^{- 2 t} + \frac{6}{5} e^{- \frac{t}{2}}