What is the solution for y^{''}+8y=0,y(0)=1,y^'(0)=1 ?

The solution for y^{''}+8y=0,y(0)=1,y^'(0)=1 is y=cos(2^{1+1/2}t)+(sin(2^{1+1/2}t))/(2sqrt(2))

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y^{''}+8y=0,y(0)=1,y^'(0)=1

y^{^{''}} + 8 y = 0, y (0) = 1, y^{^{'}} (0) = 1

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

Solution steps

y^{''} (t) + 8 y = 0

Solve linear ODE:

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

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

What is the solution for y^{''}+8y=0,y(0)=1,y^'(0)=1 ?
The solution for y^{''}+8y=0,y(0)=1,y^'(0)=1 is y=cos(2^{1+1/2}t)+(sin(2^{1+1/2}t))/(2sqrt(2))