What is the solution for 2x^{''}+32x=0,x(0)=0,x^'(0)=0 ?

The solution for 2x^{''}+32x=0,x(0)=0,x^'(0)=0 is x=0

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2x^{''}+32x=0,x(0)=0,x^'(0)=0

2 x^{^{''}} + 32 x = 0, x (0) = 0, x^{^{'}} (0) = 0

x = 0

Solution steps

2 x^{''} (t) + 32 x = 0

Solve linear ODE:

x = 0

x = 0

y^{^{''}} + y = 0, y (0) = 11, y (0) = 33

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

y^{^{'''}} + 2 y^{^{''}} - 11 y^{^{'}} - 52 y = 0

\frac{d ^{2} f}{d x ^{2}} + 64 f (x) = 0

5 y^{^{''}} - 7 y^{^{'}} + 2 y = 0, y (0) = 1, y^{^{'}} (0) = - 0.8 y

What is the solution for 2x^{''}+32x=0,x(0)=0,x^'(0)=0 ?
The solution for 2x^{''}+32x=0,x(0)=0,x^'(0)=0 is x=0