What is the solution for y^{''}+y=0,y(pi)=1,y^'(pi)=-5 ?

The solution for y^{''}+y=0,y(pi)=1,y^'(pi)=-5 is y=-cos(t)+5sin(t)

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y^{''}+y=0,y(pi)=1,y^'(pi)=-5

y^{^{''}} + y = 0, y (π) = 1, y^{^{'}} (π) = - 5

y = - cos (t) + 5 sin (t)

Solution steps

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

Solve linear ODE:

y = - cos (t) + 5 sin (t)

y = - cos (t) + 5 sin (t)

5 y^{^{''}} - 30 y^{^{'}} + 170 y = 0

y^{^{''}} + 14 y^{^{'}} + 49 y = 0, y (- 4) = 1, y^{^{'}} (- 4) = 5

y^{^{'''}} - 49 y^{^{'}} = 0

\frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} = - y

y^{^{''}} + 2 y^{^{'}} + 10 y = 0, y (0) = e^{- \frac{π}{6}}

What is the solution for y^{''}+y=0,y(pi)=1,y^'(pi)=-5 ?
The solution for y^{''}+y=0,y(pi)=1,y^'(pi)=-5 is y=-cos(t)+5sin(t)