What is the solution for y^{''}+4y=2t ?

The solution for y^{''}+4y=2t is y=c_{1}cos(2t)+c_{2}sin(2t)+t/2

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y^{''}+4y=2t

y^{^{''}} + 4 y = 2 t

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{t}{2}

Solution steps

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

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{t}{2}

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{t}{2}

y^{^{''}} + 9 y = cos (2 x)

y^{^{''}} - 10 y^{^{'}} + 25 y = x

d (x) = - 3 x^{2} - 15 x + 146

\frac{d ^{2} y}{d x ^{2}} + y = x sin (x)

\frac{1}{6} y^{^{''}} + 6 y = tan (6 t) - \frac{1}{2} e^{2 t}

What is the solution for y^{''}+4y=2t ?
The solution for y^{''}+4y=2t is y=c_{1}cos(2t)+c_{2}sin(2t)+t/2