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

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

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

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

y = c_{2} sin (t)

Solution steps

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

Solve linear ODE:

y = c_{2} sin (t)

y = c_{2} sin (t)

y = 2 x + 1 (4.3)

\int x (x - 1)^{5} d x

\frac{\partial}{\partial x} (y^{\frac{1}{3}} x^{\frac{2}{3}})

\int \frac{1}{( x - 1 ) ( x + 1 )} d x

f (x) = x^{3} - 2 x^{2} - 4 x - 6

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