What is the solution for 3y^'+2y^{''}=-cos(t) ?

The solution for 3y^'+2y^{''}=-cos(t) is y=c_{1}+c_{2}e^{-(3t)/2}-3/13 sin(t)+2/13 cos(t)

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3y^'+2y^{''}=-cos(t)

3 y^{^{'}} + 2 y^{^{''}} = - cos (t)

y = c_{1} + c_{2} e^{- \frac{3 t}{2}} - \frac{3}{13} sin (t) + \frac{2}{13} cos (t)

Solution steps

3 y^{'} (t) + 2 y^{''} (t) = - cos (t)

Solve linear ODE:

y = c_{1} + c_{2} e^{- \frac{3 t}{2}} - \frac{3}{13} sin (t) + \frac{2}{13} cos (t)

y = c_{1} + c_{2} e^{- \frac{3 t}{2}} - \frac{3}{13} sin (t) + \frac{2}{13} cos (t)

y^{^{''}} + 4 y^{^{'}} = t^{2} + 2 e^{t}, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} + 3 y = 8 sin (t)

\frac{d ^{2} y}{d t ^{2}} - y = 10 e^{t}

y^{^{''}} - 0 \cdot y^{^{'}} - 2 y = - 2 t + 4 t^{2}

y^{^{''}} - y = - 20 t + 2

What is the solution for 3y^'+2y^{''}=-cos(t) ?
The solution for 3y^'+2y^{''}=-cos(t) is y=c_{1}+c_{2}e^{-(3t)/2}-3/13 sin(t)+2/13 cos(t)