What is the solution for y^{''}+25y=0,y(pi/(10))=-3,y(pi)=-8 ?

The solution for y^{''}+25y=0,y(pi/(10))=-3,y(pi)=-8 is y=8cos(5t)-3sin(5t)

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y^{''}+25y=0,y(pi/(10))=-3,y(pi)=-8

y^{^{''}} + 25 y = 0, y (\frac{π}{10}) = - 3, y (π) = - 8

y = 8 cos (5 t) - 3 sin (5 t)

Solution steps

y^{''} (t) + 25 y = 0

Solve linear ODE:

y = 8 cos (5 t) - 3 sin (5 t)

y = 8 cos (5 t) - 3 sin (5 t)

\frac{d ^{3} y}{d x ^{3}} - 5 y = 0

10 y^{^{''}} + 4 y^{^{'}} - 2 y = 0

15 y^{^{''''}} + 68 y^{^{''}} + 64 y = 0

y^{^{''}} - 2 y^{^{'}} + 5 y = 0

y^{^{''''}} - 2 y^{^{'''}} - 4 y^{^{''}} = 0

What is the solution for y^{''}+25y=0,y(pi/(10))=-3,y(pi)=-8 ?
The solution for y^{''}+25y=0,y(pi/(10))=-3,y(pi)=-8 is y=8cos(5t)-3sin(5t)