What is the solution for y^'-1/2 y=2cos(x) ?

The solution for y^'-1/2 y=2cos(x) is y=(4(2sin(x)-cos(x)))/5+c_{1}e^{1/2 x}

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y^'-1/2 y=2cos(x)

y^{^{'}} - \frac{1}{2} y = 2 cos (x)

y = \frac{4 ( 2 sin ( x ) - cos ( x ) )}{5} + c_{1} e^{\frac{1}{2} x}

Solution steps

y^{'} (x) - \frac{1}{2} y = 2 cos (x)

Solve linear ODE:

y = \frac{4 ( 2 sin ( x ) - cos ( x ) )}{5} + c_{1} e^{\frac{1}{2} x}

y = \frac{4 ( 2 sin ( x ) - cos ( x ) )}{5} + c_{1} e^{\frac{1}{2} x}

x \to 1 lim (\frac{4}{2 x - 2})

\frac{\partial}{\partial x} (6 x^{2} y^{2} + 5 x + 8 y^{3})

\int \frac{6 x ^{3} - 5}{x} d x

x \to 1 lim (\frac{x ^{15} - 1}{x ^{5} - 1})

What is the solution for y^'-1/2 y=2cos(x) ?
The solution for y^'-1/2 y=2cos(x) is y=(4(2sin(x)-cos(x)))/5+c_{1}e^{1/2 x}