What is the solution for x^{''}+16x=0,x(0)=0,x^'(0)=5 ?

The solution for x^{''}+16x=0,x(0)=0,x^'(0)=5 is x= 5/4 sin(4t)

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x^{''}+16x=0,x(0)=0,x^'(0)=5

x^{^{''}} + 16 x = 0, x (0) = 0, x^{^{'}} (0) = 5

x = \frac{5}{4} sin (4 t)

Solution steps

x^{''} (t) + 16 x = 0

Solve linear ODE:

x = \frac{5}{4} sin (4 t)

x = \frac{5}{4} sin (4 t)

y^{^{''}} + 12 y^{^{'}} + 36 y = 0, y (0) = 3, y^{^{'}} (0) = 0

y^{^{''''}} - 42 y^{^{'''}} + 441 y^{^{''}} = 0

{\frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} + 20 y}^{2} = 0

3 y^{^{'''}} + 17 y^{^{''}} + 31 y^{^{'}} + 14 y = 0

2 y^{^{''}} - 3 y^{^{'}} + \frac{1}{2} y = 0

What is the solution for x^{''}+16x=0,x(0)=0,x^'(0)=5 ?
The solution for x^{''}+16x=0,x(0)=0,x^'(0)=5 is x= 5/4 sin(4t)