What is the solution for 1.6x^{''}+40x=0,x(0)=-1/3 ,x^'(0)= 5/4 ?

The solution for 1.6x^{''}+40x=0,x(0)=-1/3 ,x^'(0)= 5/4 is x=-1/3 cos(5t)+1/4 sin(5t)

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1.6x^{''}+40x=0,x(0)=-1/3 ,x^'(0)= 5/4

1.6 x^{^{''}} + 40 x = 0, x (0) = - \frac{1}{3}, x^{^{'}} (0) = \frac{5}{4}

x = - \frac{1}{3} cos (5 t) + \frac{1}{4} sin (5 t)

Solution steps

1.6 x^{''} (t) + 40 x = 0

Solve linear ODE:

x = - \frac{1}{3} cos (5 t) + \frac{1}{4} sin (5 t)

x = - \frac{1}{3} cos (5 t) + \frac{1}{4} sin (5 t)

y^{^{''}} - 2 a y^{^{'}} + (a^{2} - e^{2}) y = 0

y^{^{''}} - y^{^{'}} - 6 y = 0, y (0) = 2 y^{^{'}} (0) - 1

2 y^{^{''}} - 2 y^{^{'}} - 15 y = 0

4 y^{^{''}} - 784 y = 0, y (0) = 0

y^{^{'''}} + 5 y^{^{''}} - 16 y^{^{'}} - 170 y = 0

What is the solution for 1.6x^{''}+40x=0,x(0)=-1/3 ,x^'(0)= 5/4 ?
The solution for 1.6x^{''}+40x=0,x(0)=-1/3 ,x^'(0)= 5/4 is x=-1/3 cos(5t)+1/4 sin(5t)