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

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

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

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

y = 2 arctan (e^{x + c_{1}} - 1) + 2 πn - x

Solution steps

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

Solve using substitution:

y = 2 arctan (e^{x + c_{1}} - 1) + 2 πn - x

y = 2 arctan (e^{x + c_{1}} - 1) + 2 πn - x

\int (x^{2} + 2 x) e^{x} d x

3 3 x^{5} + 4 x^{4}

\frac{d}{d x} ((x^{2} + x) cos (5 x))

\int \frac{\frac{1}{x} + 1}{x ^{2}} d x

f (x) = 4 3 x, at x = 1

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