What is the solution for y^{''}-10y^'+50y=0,y(0)=1,y^'(0)=25 ?

The solution for y^{''}-10y^'+50y=0,y(0)=1,y^'(0)=25 is y=e^{5t}(cos(5t)+4sin(5t))

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y^{''}-10y^'+50y=0,y(0)=1,y^'(0)=25

y^{^{''}} - 10 y^{^{'}} + 50 y = 0, y (0) = 1, y^{^{'}} (0) = 25

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

Solution steps

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

Solve linear ODE:

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

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

What is the solution for y^{''}-10y^'+50y=0,y(0)=1,y^'(0)=25 ?
The solution for y^{''}-10y^'+50y=0,y(0)=1,y^'(0)=25 is y=e^{5t}(cos(5t)+4sin(5t))