What is the solution for 2y+sin(2t)=(dy)/(dt) ?

The solution for 2y+sin(2t)=(dy)/(dt) is y=-(cos(2t))/4-(sin(2t))/4+c_{1}e^{2t}

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2y+sin(2t)=(dy)/(dt)

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

y = - \frac{cos ( 2 t )}{4} - \frac{sin ( 2 t )}{4} + c_{1} e^{2 t}

Solution steps

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

Solve linear ODE:

y = - \frac{cos ( 2 t )}{4} - \frac{sin ( 2 t )}{4} + c_{1} e^{2 t}

y = - \frac{cos ( 2 t )}{4} - \frac{sin ( 2 t )}{4} + c_{1} e^{2 t}

\frac{d y}{d x} + y cos (x) = 4 cos (x)

\frac{\partial}{\partial x} (x + z)

\frac{d}{d x} (x^{\frac{2}{3}} y^{\frac{1}{3}})

\int (6 x - 4) d x

f (x) = arctan (\frac{1}{x})

What is the solution for 2y+sin(2t)=(dy)/(dt) ?
The solution for 2y+sin(2t)=(dy)/(dt) is y=-(cos(2t))/4-(sin(2t))/4+c_{1}e^{2t}