What is the solution for y^'+4y=0,y(0)=cos(2t),y(0)=sin(2t) ?

The solution for y^'+4y=0,y(0)=cos(2t),y(0)=sin(2t) is y=e^{-4t+c_{1}}

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y^'+4y=0,y(0)=cos(2t),y(0)=sin(2t)

y^{^{'}} + 4 y = 0, y (0) = cos (2 t), y (0) = sin (2 t)

y = e^{- 4 t + c_{1}}

Solution steps

y^{'} (t) + 4 y = 0

Solve separable ODE:

y = e^{- 4 t + c_{1}}

y = e^{- 4 t + c_{1}}

\frac{d y}{d x} = \frac{cos ( x )}{2 y}

- y = k \frac{d y}{d x}

e^{2 t} y^{^{'}} = y^{2}

y^{^{'}} - \frac{3 x ^{3}}{8 y ^{2}} = 0

\frac{d x}{d t} = \frac{( 8 x ln ( x ) )}{t}, x (1) = 4

What is the solution for y^'+4y=0,y(0)=cos(2t),y(0)=sin(2t) ?
The solution for y^'+4y=0,y(0)=cos(2t),y(0)=sin(2t) is y=e^{-4t+c_{1}}