What is the solution for y^{''}-25y^'+144y=0,y(0)=2,y^'(0)=5 ?

The solution for y^{''}-25y^'+144y=0,y(0)=2,y^'(0)=5 is y=(2-27/7)e^{16t}+27/7 e^{9t}

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y^{''}-25y^'+144y=0,y(0)=2,y^'(0)=5

y^{^{''}} - 25 y^{^{'}} + 144 y = 0, y (0) = 2, y^{^{'}} (0) = 5

y = (2 - \frac{27}{7}) e^{16 t} + \frac{27}{7} e^{9 t}

Solution steps

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

Solve linear ODE:

y = (2 - \frac{27}{7}) e^{16 t} + \frac{27}{7} e^{9 t}

y = (2 - \frac{27}{7}) e^{16 t} + \frac{27}{7} e^{9 t}

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

2 y^{^{''}} + 12 y^{^{'}} + 68 y = 0

y^{^{''}} + 14 y = 0

34 y + 10 y^{^{'}} + y^{^{''}} = 0, y (0) = - 3, y^{^{'}} (0) = 6

y^{^{''}} - 25 y = 0, y (0) = e^{5 x}

What is the solution for y^{''}-25y^'+144y=0,y(0)=2,y^'(0)=5 ?
The solution for y^{''}-25y^'+144y=0,y(0)=2,y^'(0)=5 is y=(2-27/7)e^{16t}+27/7 e^{9t}