What is the solution for y^{''}-4y^'+5y=0,y(0)=-1,y^'(0)=5 ?

The solution for y^{''}-4y^'+5y=0,y(0)=-1,y^'(0)=5 is y=e^{2t}(-cos(t)+7sin(t))

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y^{''}-4y^'+5y=0,y(0)=-1,y^'(0)=5

y^{^{''}} - 4 y^{^{'}} + 5 y = 0, y (0) = - 1, y^{^{'}} (0) = 5

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

Solution steps

y^{''} (t) - 4 y^{'} (t) + 5 y = 0

Solve linear ODE:

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

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

x, \frac{d ^{2} x}{d y ^{2}} + 0.1 x = 0

y^{^{''}} - 5 y^{^{'}} + 6 y = 0, y (0) = 1, y^{^{'}} (0) = - 1

y^{^{'''}} + 9 y^{^{''}} + 19 y^{^{'}} + 11 y = 0

y^{^{''}} = 5 y^{^{'}}

y^{^{''}} - 2 y^{^{'}} + 10 y = 0, y (0) = 1, y^{^{'}} (0) = 4

What is the solution for y^{''}-4y^'+5y=0,y(0)=-1,y^'(0)=5 ?
The solution for y^{''}-4y^'+5y=0,y(0)=-1,y^'(0)=5 is y=e^{2t}(-cos(t)+7sin(t))