What is the solution for y^{''}-25y^'+144y=0,y(0)=2 ?

The solution for y^{''}-25y^'+144y=0,y(0)=2 is y=(2-c_{2})e^{16t}+c_{2}e^{9t}

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y^{''}-25y^'+144y=0,y(0)=2

y^{^{''}} - 25 y^{^{'}} + 144 y = 0, y (0) = 2

y = (2 - c_{2}) e^{16 t} + c_{2} e^{9 t}

Solution steps

y^{''} (t) - 25 y^{'} (t) + 144 y = 0

Solve linear ODE:

y = (2 - c_{2}) e^{16 t} + c_{2} e^{9 t}

y = (2 - c_{2}) e^{16 t} + c_{2} e^{9 t}

y^{^{''}} + 2 y^{^{'}} - 8 y = 0, y^{^{'}} (0) = 14

x^{^{''}} + 2 x^{^{'}} - 8 x = 0

9 y^{^{''}} + 4 y^{^{'}} + 5 y = 0, y (0) = 2, y^{^{'}} (0) = 0

y^{^{''}} - 4 y^{^{'}} + 4 y = 0, y (0) = 5, y^{^{'}} (0) = 2

y^{^{''}} + π^{2} y = 0, y (0) = 1, y^{^{'}} (0) = 1

What is the solution for y^{''}-25y^'+144y=0,y(0)=2 ?
The solution for y^{''}-25y^'+144y=0,y(0)=2 is y=(2-c_{2})e^{16t}+c_{2}e^{9t}