What is the solution for y^{''}+1/4 y=0,y(pi)=1,y^'(pi)=-3 ?

The solution for y^{''}+1/4 y=0,y(pi)=1,y^'(pi)=-3 is y=6cos(t/2)+sin(t/2)

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y^{''}+1/4 y=0,y(pi)=1,y^'(pi)=-3

y^{^{''}} + \frac{1}{4} y = 0, y (π) = 1, y^{^{'}} (π) = - 3

y = 6 cos (\frac{t}{2}) + sin (\frac{t}{2})

Solution steps

y^{''} (t) + \frac{1}{4} y = 0

Solve linear ODE:

y = 6 cos (\frac{t}{2}) + sin (\frac{t}{2})

y = 6 cos (\frac{t}{2}) + sin (\frac{t}{2})

\frac{d}{d x} ((\frac{x - 1}{x ^{2} + x + 1})^{4})

\int sin (x) + 3 d x

t an g e n t f (x) = x^{2} - x, at x = 3

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

d er i v a t i v e lo g_{b} (x)

What is the solution for y^{''}+1/4 y=0,y(pi)=1,y^'(pi)=-3 ?
The solution for y^{''}+1/4 y=0,y(pi)=1,y^'(pi)=-3 is y=6cos(t/2)+sin(t/2)