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)

\frac{d ^{2} y}{d x ^{2}} - 16 y = x e^{x}

y^{^{''}} + y = e^{- 2 π}, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} - 6 y^{^{'}} + 8 y = x^{3} - x + 1

y^{^{''}} - y = e^{2 t} + 4 t \cdot e^{2 t} + 8 t^{2} \cdot e^{t}

(d^{2} + 9) y = sec^{3} (x)

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)