What is the solution for y^{''}+36y=0,y(0)=1,y^'(0)=0 ?

The solution for y^{''}+36y=0,y(0)=1,y^'(0)=0 is y=cos(6t)

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y^{''}+36y=0,y(0)=1,y^'(0)=0

y^{^{''}} + 36 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y = cos (6 t)

Solution steps

y^{''} (t) + 36 y = 0

Solve linear ODE:

y = cos (6 t)

y = cos (6 t)

f (x) = 4 x

\frac{d}{d x} (\frac{x ^{3}}{x ^{3} + 1})

x \to 0 lim (4 x^{2} - 2 x + 3)

\int \frac{x}{1 0 ^{x}} d x

10 - 7 x, x^{2} - 20, 0

What is the solution for y^{''}+36y=0,y(0)=1,y^'(0)=0 ?
The solution for y^{''}+36y=0,y(0)=1,y^'(0)=0 is y=cos(6t)