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

The solution for y^{''}+25y=0,y(pi)=0,y(0)=0 is y=c_{2}sin(5t)

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

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

y = c_{2} sin (5 t)

Solution steps

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

Solve linear ODE:

y = c_{2} sin (5 t)

y = c_{2} sin (5 t)

y^{^{''}} = a y^{^{'}}

y^{^{''}} - 10 y^{^{'}} + 25 y = 0, y (0) = x e^{5 x}

9 y^{^{''}} - 12 y^{^{'}} + 4 y = 0, y (0) = 2, y^{^{'}} (0) = 1

\frac{d ^{2} y}{d x ^{2}} - 10 \frac{d y}{d x} + 29 y = 0

y^{^{''}} + 10 y^{^{'}} + 16 y = 0, y (0) = 1, y^{^{'}} (0) = 0

What is the solution for y^{''}+25y=0,y(pi)=0,y(0)=0 ?
The solution for y^{''}+25y=0,y(pi)=0,y(0)=0 is y=c_{2}sin(5t)