What is the solution to the Differential Equation #dy/dx = sin(x+y) + cos(x+y)#?

1 Answer
Oct 30, 2016

# -log(abs(2*sin(y)/(cos(y)+1)-2^(3/2)-2)/abs(2*sin(y)/(cos(y)+1)+2^(3/2)-2))/sqrt(2) = -cosx + sinx + C#

Explanation:

# dy/dx = sin(x+y) + cos(x+y) #

In it's present form this is not separable, but using the sine and cosine sum formula we have:

# dy/dx = sinxcosy+cosxsiny + cosxcosy+sinxsiny #
# :. dy/dx = sinxcosy+sinxsiny+cosxsiny + cosxcosy #
# :. dy/dx = sinx(cosy+siny)+cosx(siny + cosy) #
# :. dy/dx = sinx(siny+cosy)+cosx(siny + cosy) #
# :. dy/dx = (siny + cosy)(sinx + cosx) #

This is now separable (phew! But you do need to be good with trigonometry!), So "separating the variables" gives us:

# int 1/((siny + cosy))dy = int (sinx + cosx) dx#

The LHS integral is a real pig, This question is more about the separation process than horrific integration, so I will just quote the result which is:

#int 1/((siny + cosy))dy = -log(abs(2*sin(y)/(cos(y)+1)-2^(3/2)-2)/abs(2*sin(y)/(cos(y)+1)+2^(3/2)-2))/sqrt(2)#

So the solution is:
# -log(abs(2*sin(y)/(cos(y)+1)-2^(3/2)-2)/abs(2*sin(y)/(cos(y)+1)+2^(3/2)-2))/sqrt(2) = -cosx + sinx + C#