What is the solution for y^{''}-9y^'+20y=0,y(0)=1,y^'(0)=2 ?

The solution for y^{''}-9y^'+20y=0,y(0)=1,y^'(0)=2 is y=-2e^{5t}+3e^{4t}

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y^{''}-9y^'+20y=0,y(0)=1,y^'(0)=2

y^{^{''}} - 9 y^{^{'}} + 20 y = 0, y (0) = 1, y^{^{'}} (0) = 2

y = - 2 e^{5 t} + 3 e^{4 t}

Solution steps

y^{''} (t) - 9 y^{'} (t) + 20 y = 0

Solve linear ODE:

y = - 2 e^{5 t} + 3 e^{4 t}

y = - 2 e^{5 t} + 3 e^{4 t}

u^{^{''}} + 2 u^{^{'}} + 2 H (t) = 0

55 x^{^{''}} + 220 x = 0, x (0) = 0, x^{^{'}} (0) = - 6

y^{^{'''}} + 3.1 y^{^{''}} + 3.1 y^{^{'}} + y = 0

y^{^{''}} + 6 y^{^{'}} + 8 y = 0, y (0) = \frac{2}{5}, y^{^{'}} (0) = \frac{2}{5}

y^{^{''}} + 20 y^{^{'}} + 36 y = 0, y^{^{'}} (0) = 0, y (0) = 5

What is the solution for y^{''}-9y^'+20y=0,y(0)=1,y^'(0)=2 ?
The solution for y^{''}-9y^'+20y=0,y(0)=1,y^'(0)=2 is y=-2e^{5t}+3e^{4t}