What is the solution for y^'-7y=sin(2x) ?

The solution for y^'-7y=sin(2x) is y=-(2cos(2x))/(53)-(7sin(2x))/(53)+c_{1}e^{7x}

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y^'-7y=sin(2x)

y^{^{'}} - 7 y = sin (2 x)

y = - \frac{2 cos ( 2 x )}{53} - \frac{7 sin ( 2 x )}{53} + c_{1} e^{7 x}

Solution steps

y^{'} (x) - 7 y = sin (2 x)

Solve linear ODE:

y = - \frac{2 cos ( 2 x )}{53} - \frac{7 sin ( 2 x )}{53} + c_{1} e^{7 x}

y = - \frac{2 cos ( 2 x )}{53} - \frac{7 sin ( 2 x )}{53} + c_{1} e^{7 x}

y = 1 + x - 4 x^{\frac{1}{2}}

arctan (\frac{1 + x}{1 - x})

2 + \frac{d y}{d x} = 2 x + y

\frac{d}{d x} (\frac{in x}{x})

x \to 8 lim (e^{\frac{3}{( x - 8 )}})

What is the solution for y^'-7y=sin(2x) ?
The solution for y^'-7y=sin(2x) is y=-(2cos(2x))/(53)-(7sin(2x))/(53)+c_{1}e^{7x}