What is the solution for y^{''}-23y^'+112y=0,y(0)=3,y^'(0)=48 ?

The solution for y^{''}-23y^'+112y=0,y(0)=3,y^'(0)=48 is y=3e^{16t}

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y^{''}-23y^'+112y=0,y(0)=3,y^'(0)=48

y^{^{''}} - 23 y^{^{'}} + 112 y = 0, y (0) = 3, y^{^{'}} (0) = 48

y = 3 e^{16 t}

Solution steps

y^{''} (t) - 23 y^{'} (t) + 112 y = 0

Solve linear ODE:

y = 3 e^{16 t}

y = 3 e^{16 t}

y^{^{'''}} + 8 y^{^{''}} = 0

f^{^{''}} (t) + f (t) = 0, f^{^{'}} (0) = 1

y^{^{''''}} + 5 y^{^{'''}} - 5 y^{^{''}} - 5 y^{^{'}} + \frac{1}{2} y = 0

2 x^{^{''}} + 7 x = 0

(D^{2} + 4 D + 29)^{2} y = 0

What is the solution for y^{''}-23y^'+112y=0,y(0)=3,y^'(0)=48 ?
The solution for y^{''}-23y^'+112y=0,y(0)=3,y^'(0)=48 is y=3e^{16t}