What is the solution for y^{''}-20y^'+101y=0,y(0)=8,y^'(0)=1 ?

The solution for y^{''}-20y^'+101y=0,y(0)=8,y^'(0)=1 is y=e^{10t}(8cos(t)-79sin(t))

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y^{''}-20y^'+101y=0,y(0)=8,y^'(0)=1

y^{^{''}} - 20 y^{^{'}} + 101 y = 0, y (0) = 8, y^{^{'}} (0) = 1

y = e^{10 t} (8 cos (t) - 79 sin (t))

Solution steps

y^{''} (t) - 20 y^{'} (t) + 101 y = 0

Solve linear ODE:

y = e^{10 t} (8 cos (t) - 79 sin (t))

y = e^{10 t} (8 cos (t) - 79 sin (t))

r^{2} y^{^{''}} + r y^{^{'}} - n^{2} y = 0

\frac{1}{4} y^{^{''}} + \frac{7}{4} y^{^{'}} + \frac{5}{2} y = 0

y^{^{''}} - 8 y^{^{'}} + 16 y = 0, y (0) = 0, y^{^{'}} (0) = 3

y^{^{''}} - 8 y^{^{'}} + 16 y = 0, y (0) = 0, y^{^{'}} (0) = 8

\frac{d ^{2} y}{d x ^{2}} + 9 \frac{d y}{d x} + 9 y = 0

What is the solution for y^{''}-20y^'+101y=0,y(0)=8,y^'(0)=1 ?
The solution for y^{''}-20y^'+101y=0,y(0)=8,y^'(0)=1 is y=e^{10t}(8cos(t)-79sin(t))