What is the solution for y^'-8y=cos(x) ?

The solution for y^'-8y=cos(x) is y=(sin(x))/(65)-(8cos(x))/(65)+c_{1}e^{8x}

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y^'-8y=cos(x)

y^{^{'}} - 8 y = cos (x)

y = \frac{sin ( x )}{65} - \frac{8 cos ( x )}{65} + c_{1} e^{8 x}

Solution steps

y^{'} (x) - 8 y = cos (x)

Solve linear ODE:

y = \frac{sin ( x )}{65} - \frac{8 cos ( x )}{65} + c_{1} e^{8 x}

y = \frac{sin ( x )}{65} - \frac{8 cos ( x )}{65} + c_{1} e^{8 x}

What is the solution for y^'-8y=cos(x) ?
The solution for y^'-8y=cos(x) is y=(sin(x))/(65)-(8cos(x))/(65)+c_{1}e^{8x}