What is the solution for y^'-y=2,cos(5t)t(t)=0 ?

The solution for y^'-y=2,cos(5t)t(t)=0 is y=e^{t+c_{1}}-2

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y^'-y=2,cos(5t)t(t)=0

y^{^{'}} - y = 2, cos (5 t) t (t) = 0

y = e^{t + c_{1}} - 2

Solution steps

y^{'} (t) - y = 2

Solve separable ODE:

y = e^{t + c_{1}} - 2

y = e^{t + c_{1}} - 2

y^{^{'}} = 5 y + 4

2 x y + \frac{d y}{d x} = \frac{6 x}{y ^{2}}

(2 x y - y - 4 x + 2) d x + d y = 0

\frac{1 + cos ( 2 x )}{1 + sin ( y )} + y^{^{'}} = 0

e^{- x} sin (y) \frac{d y}{d x} = 1 + e^{- x}

What is the solution for y^'-y=2,cos(5t)t(t)=0 ?
The solution for y^'-y=2,cos(5t)t(t)=0 is y=e^{t+c_{1}}-2