What is the solution for y^'+ycos(x)=3cos(x),y(0)=5 ?

The solution for y^'+ycos(x)=3cos(x),y(0)=5 is y=2e^{-sin(x)}+3

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y^'+ycos(x)=3cos(x),y(0)=5

y^{^{'}} + y cos (x) = 3 cos (x), y (0) = 5

y = 2 e^{- s i n (x)} + 3

Solution steps

y^{'} (x) + y cos (x) = 3 cos (x)

Solve separable ODE:

y = 2 e^{- s i n (x)} + 3

y = 2 e^{- s i n (x)} + 3

x \frac{d y}{d x} - 7 y = 0

e^{y} y^{^{'}} = \frac{( ln ( t ) ) ^{2}}{t}, y (1) = ln (19)

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

y^{^{'}} = x (y^{2} + 1)

What is the solution for y^'+ycos(x)=3cos(x),y(0)=5 ?
The solution for y^'+ycos(x)=3cos(x),y(0)=5 is y=2e^{-sin(x)}+3