What is the solution for y^{''}+4y=t,y(0)=0,y^'(0)=0 ?

The solution for y^{''}+4y=t,y(0)=0,y^'(0)=0 is y= t/4-1/8 sin(2t)

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y^{''}+4y=t,y(0)=0,y^'(0)=0

y^{^{''}} + 4 y = t, y (0) = 0, y^{^{'}} (0) = 0

y = \frac{t}{4} - \frac{1}{8} sin (2 t)

Solution steps

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

Solve linear ODE:

y = \frac{t}{4} - \frac{1}{8} sin (2 t)

y = \frac{t}{4} - \frac{1}{8} sin (2 t)

What is the solution for y^{''}+4y=t,y(0)=0,y^'(0)=0 ?
The solution for y^{''}+4y=t,y(0)=0,y^'(0)=0 is y= t/4-1/8 sin(2t)