What is the solution for y^{''}+4y=sin(x),y(0)=0,y(1)=0 ?

The solution for y^{''}+4y=sin(x),y(0)=0,y(1)=0 is y=-(sin(1)sin(2x))/(3sin(2))+1/3 sin(x)

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y^{''}+4y=sin(x),y(0)=0,y(1)=0

y^{^{''}} + 4 y = sin (x), y (0) = 0, y (1) = 0

y = - \frac{sin ( 1 ) sin ( 2 x )}{3 sin ( 2 )} + \frac{1}{3} sin (x)

Solution steps

y^{''} (x) + 4 y = sin (x)

Solve linear ODE:

y = - \frac{sin ( 1 ) sin ( 2 x )}{3 sin ( 2 )} + \frac{1}{3} sin (x)

y = - \frac{sin ( 1 ) sin ( 2 x )}{3 sin ( 2 )} + \frac{1}{3} sin (x)

What is the solution for y^{''}+4y=sin(x),y(0)=0,y(1)=0 ?
The solution for y^{''}+4y=sin(x),y(0)=0,y(1)=0 is y=-(sin(1)sin(2x))/(3sin(2))+1/3 sin(x)