What is the solution for y^{''}-10y^'+50y=0,y(0)=-4,y^'(0)=-5 ?

The solution for y^{''}-10y^'+50y=0,y(0)=-4,y^'(0)=-5 is y=e^{5t}(-4cos(5t)+3sin(5t))

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y^{''}-10y^'+50y=0,y(0)=-4,y^'(0)=-5

y^{^{''}} - 10 y^{^{'}} + 50 y = 0, y (0) = - 4, y^{^{'}} (0) = - 5

y = e^{5 t} (- 4 cos (5 t) + 3 sin (5 t))

Solution steps

y^{''} (t) - 10 y^{'} (t) + 50 y = 0

Solve linear ODE:

y = e^{5 t} (- 4 cos (5 t) + 3 sin (5 t))

y = e^{5 t} (- 4 cos (5 t) + 3 sin (5 t))

9 y^{^{''}} - 12 y^{^{'}} + 4 y = 0, y (0) = e^{(\frac{2 x}{3})}

y^{^{''}} + 2 y^{^{'}} + (1 + \frac{9}{4} π^{2}) y = 0

y^{^{''}} + 16 y = 0, y (0) = cos (4 x)

y^{^{''}} - y^{^{'}} - y = 0, y^{1} (0) = e^{x}, y^{2} (0) = x

12 y^{^{''}} - 5 y^{^{'}} - 3 y = 0

What is the solution for y^{''}-10y^'+50y=0,y(0)=-4,y^'(0)=-5 ?
The solution for y^{''}-10y^'+50y=0,y(0)=-4,y^'(0)=-5 is y=e^{5t}(-4cos(5t)+3sin(5t))