What is the solution for y^'+(cos(x))y=5cos(x) ?

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

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y^'+(cos(x))y=5cos(x)

y^{^{'}} + (cos (x)) y = 5 cos (x)

y = e^{- s i n (x) + c_{1}} + 5

Solution steps

y^{'} (x) + (cos (x)) y = 5 cos (x)

Solve separable ODE:

y = e^{- s i n (x) + c_{1}} + 5

y = e^{- s i n (x) + c_{1}} + 5

y^{^{'}} = \frac{3}{2} y^{\frac{1}{3}}, y (0) = 0

x^{^{'}} = \frac{( x ^{2} + x )}{( 2 x + 1 )}

3 = \frac{π h ^{2}}{4} \frac{d h}{d t}

sin (x) \frac{d y}{d x} + 3 y = x^{3} y

e^{- 2 x + 4 y} d x + e^{3 x - 6 y} d y = 0

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