Question

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

Answer 1

`-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`