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

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

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

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

y = \frac{e ^{16 t} ( 5 - 9 c _{2} )}{16} + c_{2} e^{9 t}

Solution steps

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

Solve linear ODE:

y = \frac{e ^{16 t} ( 5 - 9 c _{2} )}{16} + c_{2} e^{9 t}

y = \frac{e ^{16 t} ( 5 - 9 c _{2} )}{16} + c_{2} e^{9 t}

y^{^{''}} - 12 y^{^{'}} + 36 y = 0, y (0) = 4, y^{^{'}} (0) = 15.455

3 y^{^{''}} + 4 y^{^{'}} + 5 y = 0

y^{^{''}} + \frac{49}{3} y = 0

y^{^{''}} - 5 y^{^{''}} + 24 y^{^{'}} - 20 y = 0

y^{^{''}} - 12 y^{^{'}} + 36 y = 0, y (0) = 4, y^{^{'}} (0) = 15.449

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