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

The solution for y^'+4y=cos(4t),y(0)=2 is y=(sin(4t))/8+(cos(4t))/8+(15)/(8e^{4t)}

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

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

y = \frac{sin ( 4 t )}{8} + \frac{cos ( 4 t )}{8} + \frac{15}{8 e ^{4 t}}

Solution steps

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

Solve linear ODE:

y = \frac{sin ( 4 t )}{8} + \frac{cos ( 4 t )}{8} + \frac{15}{8 e ^{4 t}}

y = \frac{sin ( 4 t )}{8} + \frac{cos ( 4 t )}{8} + \frac{15}{8 e ^{4 t}}

\frac{1}{3} x^{\frac{3}{2}} - x^{\frac{1}{2}}

\int \frac{1}{x ^{3}} d x

f (x) = x^{\frac{1}{5}}

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

x \to 2 lim (5 x - 14 x^{- 2})

What is the solution for y^'+4y=cos(4t),y(0)=2 ?
The solution for y^'+4y=cos(4t),y(0)=2 is y=(sin(4t))/8+(cos(4t))/8+(15)/(8e^{4t)}