The general solution of the differential equation dy/dx=tan(x+y) is equal to y= x -log|cos(x+y) + sin(x+y)| +C where C denotes an arbitrary constant. In this post, we will learn how to solve the differential equation dy/dx = tan(x+y).

## General Solution of dy/dx=tan(x+y)

**Question:** Find the general solution of $\dfrac{dy}{dx}$=tan(x+y).

**Solution:**

$\dfrac{dy}{dx}$=tan(x+y)

⇒ $\dfrac{dy}{dx} = \dfrac{\sin(x+y)}{\cos(x+y)}$ .**..(∗)**

Put x+y = v

So, 1+ $\dfrac{dy}{dx} = \dfrac{dv}{dx}$

⇒ $\dfrac{dy}{dx} = \dfrac{dv}{dx} -1$

Therefore, from (∗), we get that

$\dfrac{dv}{dx}-1 = \dfrac{\sin v}{\cos v}$

⇒ $\dfrac{dv}{dx} = \dfrac{\cos v +\sin v}{\cos v}$

⇒ $\dfrac{\cos v}{\cos v +\sin v} dv = dx$

Integrating, $\dfrac{1}{2}\int \dfrac{(\cos v +\sin v) +(\cos v -\sin v)}{\cos v +\sin v} dv = \int dx +K$

⇒ $\dfrac{1}{2} \int dv + \dfrac{1}{2} \int \dfrac{\frac{d}{dv}(\cos v +\sin v)}{\cos v +\sin v} dv = x +K$

⇒ $\dfrac{v}{2}+ \dfrac{1}{2} \log |\cos v +\sin v| = x +K$

⇒ $v+ \log |\cos v +\sin v| = 2x +C$ where C=2K.

⇒ $x+y+\log |\cos (x+y) +\sin (x+y)| = 2x +C$ as v=x+y.

⇒ $y=x-\log |\cos (x+y) +\sin (x+y)|+C$

So the solution of dy/dx=tan(x+y) is equal to y= x -log|cos(x+y) + sin(x+y)| +C where C is an integral constant.

**More Differential Equations:**

## FAQs

### Q1: What is the general solution of dy/dx =tan(x+y)?

**Answer:** The general solution of dy/dx=tan(x+y) is given by y= x -log|cos(x+y) + sin(x+y)| +C where C is a constant.