What is the solution for y^{''}-2y^'+5y=0,y(0)=-4,y^'(0)=3 ?

The solution for y^{''}-2y^'+5y=0,y(0)=-4,y^'(0)=3 is y=e^t(-4cos(2t)+7/2 sin(2t))

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y^{''}-2y^'+5y=0,y(0)=-4,y^'(0)=3

y^{^{''}} - 2 y^{^{'}} + 5 y = 0, y (0) = - 4, y^{^{'}} (0) = 3

y = e^{t} (- 4 cos (2 t) + \frac{7}{2} sin (2 t))

Solution steps

y^{''} (t) - 2 y^{'} (t) + 5 y = 0

Solve linear ODE:

y = e^{t} (- 4 cos (2 t) + \frac{7}{2} sin (2 t))

y = e^{t} (- 4 cos (2 t) + \frac{7}{2} sin (2 t))

x^{^{''}} + 4 x^{^{'}} - 5 x = 0

9 y^{^{''}} - 42 y^{^{'}} + 49 y = 0

x^{^{''}} + 2 a x^{^{'}} + b x = 0

y^{^{''''}} + 4 y^{^{'''}} + 4 y^{^{''}} - 4 y^{^{'}} - 5 y = 0

x^{^{''}} - 6 x^{^{''}} + x = 0

What is the solution for y^{''}-2y^'+5y=0,y(0)=-4,y^'(0)=3 ?
The solution for y^{''}-2y^'+5y=0,y(0)=-4,y^'(0)=3 is y=e^t(-4cos(2t)+7/2 sin(2t))