What is the solution for y^'=2y+sin(2t) ?

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

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y^'=2y+sin(2t)

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

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

Solution steps

y^{'} (t) = 2 y + sin (2 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{\partial}{\partial x} (e^{- (x^{3} + y^{3})})

\frac{\partial}{\partial x} (x^{3} + y^{3} - 5 x y)

\frac{d}{d x} (\frac{4 + x}{1 - 4 x})

f (x) = x^{3} - 5 x + 3, at x = 3

f (x) = (- x^{2} - 4)^{2}

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