What is the solution for x^{''}+16x=0,x(0)=-1,x^'(0)=-6 ?

The solution for x^{''}+16x=0,x(0)=-1,x^'(0)=-6 is x=-cos(4t)-3/2 sin(4t)

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x^{''}+16x=0,x(0)=-1,x^'(0)=-6

x^{^{''}} + 16 x = 0, x (0) = - 1, x^{^{'}} (0) = - 6

x = - cos (4 t) - \frac{3}{2} sin (4 t)

Solution steps

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

Solve linear ODE:

x = - cos (4 t) - \frac{3}{2} sin (4 t)

x = - cos (4 t) - \frac{3}{2} sin (4 t)

4 y^{^{''}} + 20 y^{^{'}} + 25 y = 0, y (0) = 1, y^{^{'}} (0) = 2

(D^{3} - 3 D^{2} + 4) y = 0

y^{^{''}} - 24 y^{^{'}} + α y = 0, y (0) = 1, y^{^{'}} (0) = 22

\frac{d ^{2} y}{d x ^{2}} - a y = 0

6 \frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} - y = 0

What is the solution for x^{''}+16x=0,x(0)=-1,x^'(0)=-6 ?
The solution for x^{''}+16x=0,x(0)=-1,x^'(0)=-6 is x=-cos(4t)-3/2 sin(4t)