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

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

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

y^{^{''}} + 25 y = 0, y (π) = 3, y^{^{'}} (π) = - 3

y = - 3 cos (5 t) + \frac{3}{5} sin (5 t)

Solution steps

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

Solve linear ODE:

y = - 3 cos (5 t) + \frac{3}{5} sin (5 t)

y = - 3 cos (5 t) + \frac{3}{5} sin (5 t)

\int_{0}^{4} (- x^{2} + 4 x) d x

(1) (- 1.5)

y^{^{'}} + 3 y = 2 e^{- x}

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

\int 9 t sin (t) cos (t) d t

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