What is the solution for y^{''}+4y=4sin(2t),y(pi)=0,y^'(pi)=-1 ?

The solution for y^{''}+4y=4sin(2t),y(pi)=0,y^'(pi)=-1 is y=-tcos(2t)+picos(2t)

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y^{''}+4y=4sin(2t),y(pi)=0,y^'(pi)=-1

y^{^{''}} + 4 y = 4 sin (2 t), y (π) = 0, y^{^{'}} (π) = - 1

y = - t cos (2 t) + π cos (2 t)

Solution steps

y^{''} (t) + 4 y = 4 sin (2 t)

Solve linear ODE:

y = - t cos (2 t) + π cos (2 t)

y = - t cos (2 t) + π cos (2 t)

2 y^{^{''}} + 8 y^{^{'}} + 80 y = 40, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 16 y = 12 cos (4 x) + 18 sin (4 x)

- \frac{1}{4} y^{^{''}} + y^{^{'}} + y = x^{2} - 2 x

y^{^{''}} - 10 y^{^{'}} + 25 y = \frac{e ^{5 x}}{4 + x ^{2}}

y^{^{''}} - 9 y = sin (3 x)

What is the solution for y^{''}+4y=4sin(2t),y(pi)=0,y^'(pi)=-1 ?
The solution for y^{''}+4y=4sin(2t),y(pi)=0,y^'(pi)=-1 is y=-tcos(2t)+picos(2t)