What is the solution for x^{''}+9x=0,x(0)=1 ?

The solution for x^{''}+9x=0,x(0)=1 is x=cos(3t)+c_{2}sin(3t)

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x^{''}+9x=0,x(0)=1

x^{^{''}} + 9 x = 0, x (0) = 1

x = cos (3 t) + c_{2} sin (3 t)

Solution steps

x^{''} (t) + 9 x = 0

Solve linear ODE:

x = cos (3 t) + c_{2} sin (3 t)

x = cos (3 t) + c_{2} sin (3 t)

y^{^{''''}} - 4 y^{^{'''}} + 20 y^{^{''}} - 64 y^{^{'}} + 64 y = 0

\frac{d ^{2} x}{d t ^{2}} + 4 \frac{d x}{d t} + 5 x = 0, x (0) = 1

9 y^{^{''}} + 8 y^{^{'}} + 7 y = 0

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

\frac{d ^{2} y}{d t ^{2}} + 7 \frac{d y}{d t} + 12 y = 0

What is the solution for x^{''}+9x=0,x(0)=1 ?
The solution for x^{''}+9x=0,x(0)=1 is x=cos(3t)+c_{2}sin(3t)