What is the solution for y^{''}+y^'+5/4 y=0,y(0)=3,y^'(0)=1 ?

The solution for y^{''}+y^'+5/4 y=0,y(0)=3,y^'(0)=1 is y=e^{-t/2}(3cos(t)+5/2 sin(t))

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y^{''}+y^'+5/4 y=0,y(0)=3,y^'(0)=1

y^{^{''}} + y^{^{'}} + \frac{5}{4} y = 0, y (0) = 3, y^{^{'}} (0) = 1

y = e^{- \frac{t}{2}} (3 cos (t) + \frac{5}{2} sin (t))

Solution steps

y^{''} (t) + y^{'} (t) + \frac{5}{4} y = 0

Solve linear ODE:

y = e^{- \frac{t}{2}} (3 cos (t) + \frac{5}{2} sin (t))

y = e^{- \frac{t}{2}} (3 cos (t) + \frac{5}{2} sin (t))

y^{^{''}} - 14 y^{^{'}} + 58 y = 0

y^{^{''}} + 2 y^{^{'}} + y = 0, y^{^{'}} (0) = 8, y (1) = 8

y^{^{''}} - 2 y^{^{'}} + 10 y = 0, y (0) = 0, y^{^{'}} (0) = 0

\frac{d ^{2} x}{d t ^{2}} + 2 \frac{d x}{d t} + 4 x = 0

2 y^{^{''}} + y^{^{'}} + 2 y = 0

What is the solution for y^{''}+y^'+5/4 y=0,y(0)=3,y^'(0)=1 ?
The solution for y^{''}+y^'+5/4 y=0,y(0)=3,y^'(0)=1 is y=e^{-t/2}(3cos(t)+5/2 sin(t))