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

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

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

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

y = \frac{1}{6} sin (6 t)

Solution steps

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

Solve linear ODE:

y = \frac{1}{6} sin (6 t)

y = \frac{1}{6} sin (6 t)

2 - 3 x

\int 4 - y^{2} d y

n = 1 \sum \infty \frac{- 1 ^{n}}{n ^{5}}

\int \frac{2 ( 1 + x )}{x ( 2 + x )} d x

\int_{0}^{\infty} e^{- \frac{x}{3}} d x

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