What is the solution for y^{''}+4y=0,y(0)=0,y^'(0)=1 ?

The solution for y^{''}+4y=0,y(0)=0,y^'(0)=1 is y= 1/2 sin(2t)

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

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

y = \frac{1}{2} sin (2 t)

Solution steps

y^{''} (t) + 4 y = 0

Solve linear ODE:

y = \frac{1}{2} sin (2 t)

y = \frac{1}{2} sin (2 t)

\frac{d y}{d x} = \frac{3}{2 x ^{5}}

\int \frac{1}{- x ^{2} + 2 x + 3} d x

\int \frac{7}{x ( x ^{4} + 8 )} d x

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

\frac{3}{4 x + 1}

What is the solution for y^{''}+4y=0,y(0)=0,y^'(0)=1 ?
The solution for y^{''}+4y=0,y(0)=0,y^'(0)=1 is y= 1/2 sin(2t)