What is the solution for y^'+7y=5cos(2t) ?

The solution for y^'+7y=5cos(2t) is y=(5(2sin(2t)+7cos(2t)))/(53)+(c_{1})/(e^{7t)}

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y^'+7y=5cos(2t)

y^{^{'}} + 7 y = 5 cos (2 t)

y = \frac{5 ( 2 sin ( 2 t ) + 7 cos ( 2 t ) )}{53} + \frac{c _{1}}{e ^{7 t}}

Solution steps

y^{'} (t) + 7 y = 5 cos (2 t)

Solve linear ODE:

y = \frac{5 ( 2 sin ( 2 t ) + 7 cos ( 2 t ) )}{53} + \frac{c _{1}}{e ^{7 t}}

y = \frac{5 ( 2 sin ( 2 t ) + 7 cos ( 2 t ) )}{53} + \frac{c _{1}}{e ^{7 t}}

y^{^{''}} - 2 y^{^{'}} + 2 y = 2 x - 2

x \to 0 + lim (2^{- \frac{1}{x}})

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\int \frac{4 x ^{2} + x + 27}{x ^{3} + 9 x} d x

\frac{\partial}{\partial x} (2 x^{2} - 7 y^{9} - 4)

What is the solution for y^'+7y=5cos(2t) ?
The solution for y^'+7y=5cos(2t) is y=(5(2sin(2t)+7cos(2t)))/(53)+(c_{1})/(e^{7t)}