What is the solution for y^{''}-2y^'+10y=0,y(0)=0,y^'(0)=-4 ?

The solution for y^{''}-2y^'+10y=0,y(0)=0,y^'(0)=-4 is y=-4/3 e^tsin(3t)

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y^{''}-2y^'+10y=0,y(0)=0,y^'(0)=-4

y^{^{''}} - 2 y^{^{'}} + 10 y = 0, y (0) = 0, y^{^{'}} (0) = - 4

y = - \frac{4}{3} e^{t} sin (3 t)

Solution steps

y^{''} (t) - 2 y^{'} (t) + 10 y = 0

Solve linear ODE:

y = - \frac{4}{3} e^{t} sin (3 t)

y = - \frac{4}{3} e^{t} sin (3 t)

What is the solution for y^{''}-2y^'+10y=0,y(0)=0,y^'(0)=-4 ?
The solution for y^{''}-2y^'+10y=0,y(0)=0,y^'(0)=-4 is y=-4/3 e^tsin(3t)