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

Oct 30, 2016

$- \log \frac{\frac{\left\mid 2 \cdot \sin \frac{y}{\cos \left(y\right) + 1} - {2}^{\frac{3}{2}} - 2 \right\mid}{\left\mid 2 \cdot \sin \frac{y}{\cos \left(y\right) + 1} + {2}^{\frac{3}{2}} - 2 \right\mid}}{\sqrt{2}} = - \cos x + \sin x + C$

#### Explanation:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \sin \left(x + y\right) + \cos \left(x + y\right)$

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \sin x \cos y + \cos x \sin y + \cos x \cos y + \sin x \sin y$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \sin x \cos y + \sin x \sin y + \cos x \sin y + \cos x \cos y$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \sin x \left(\cos y + \sin y\right) + \cos x \left(\sin y + \cos y\right)$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \sin x \left(\sin y + \cos y\right) + \cos x \left(\sin y + \cos y\right)$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \left(\sin y + \cos y\right) \left(\sin x + \cos x\right)$

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

$\int \frac{1}{\left(\sin y + \cos y\right)} \mathrm{dy} = \int \left(\sin x + \cos x\right) \mathrm{dx}$

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:

$\int \frac{1}{\left(\sin y + \cos y\right)} \mathrm{dy} = - \log \frac{\frac{\left\mid 2 \cdot \sin \frac{y}{\cos \left(y\right) + 1} - {2}^{\frac{3}{2}} - 2 \right\mid}{\left\mid 2 \cdot \sin \frac{y}{\cos \left(y\right) + 1} + {2}^{\frac{3}{2}} - 2 \right\mid}}{\sqrt{2}}$

So the solution is:
$- \log \frac{\frac{\left\mid 2 \cdot \sin \frac{y}{\cos \left(y\right) + 1} - {2}^{\frac{3}{2}} - 2 \right\mid}{\left\mid 2 \cdot \sin \frac{y}{\cos \left(y\right) + 1} + {2}^{\frac{3}{2}} - 2 \right\mid}}{\sqrt{2}} = - \cos x + \sin x + C$