What is the solution for y^{''}+y^'+1.25y=0,y(0)=3,y^'(0)=1 ?

The solution for y^{''}+y^'+1.25y=0,y(0)=3,y^'(0)=1 is y=e^{-t/2}(3cos(t)+5/2 sin(t))

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y^{''}+y^'+1.25y=0,y(0)=3,y^'(0)=1

y^{^{''}} + y^{^{'}} + 1.25 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) + 1.25 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))

\int 23 sec^{- 2} (x t a) n^{3} x d x

\int \frac{x \cdot e ^{x}}{( x + 1 ) ^{2}} d x

\frac{d}{d x} (\frac{x ^{4} + 1}{x ^{2} + 1})

\frac{\partial}{\partial x} (y (e^{x^{2} - x y}))

What is the solution for y^{''}+y^'+1.25y=0,y(0)=3,y^'(0)=1 ?
The solution for y^{''}+y^'+1.25y=0,y(0)=3,y^'(0)=1 is y=e^{-t/2}(3cos(t)+5/2 sin(t))