What is the solution for y^{''}-9y^'+20y=0,y(0)=3,y^'(0)=2 ?

The solution for y^{''}-9y^'+20y=0,y(0)=3,y^'(0)=2 is y=-10e^{5t}+13e^{4t}

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y^{''}-9y^'+20y=0,y(0)=3,y^'(0)=2

y^{^{''}} - 9 y^{^{'}} + 20 y = 0, y (0) = 3, y^{^{'}} (0) = 2

y = - 10 e^{5 t} + 13 e^{4 t}

Solution steps

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

Solve linear ODE:

y = - 10 e^{5 t} + 13 e^{4 t}

y = - 10 e^{5 t} + 13 e^{4 t}

What is the solution for y^{''}-9y^'+20y=0,y(0)=3,y^'(0)=2 ?
The solution for y^{''}-9y^'+20y=0,y(0)=3,y^'(0)=2 is y=-10e^{5t}+13e^{4t}