The general solution of the differential equation dy/dx =sec(x+y) is equal to y =tan (x+y)/2 +C where C denotes an arbitrary constant. In this post, we will learn how to find the general solution of dy/dx=sec(x+y).

## Solve dy/dx=sec(x+y)

**Question:** What is the general solution of $\dfrac{dy}{dx}$ =sec(x+y)?

**Answer:**

Put x+y=v, so that $1+\dfrac{dy}{dx}=\dfrac{dv}{dx}$.

⇒ $\dfrac{dy}{dx}=\dfrac{dv}{dx}-1$

So, the given differential equation becomes

$\dfrac{dv}{dx}-1 =\sec v$

⇒ $\dfrac{dv}{dx} =1+\dfrac{1}{\cos v}$

⇒ $\dfrac{dv}{dx} =\dfrac{1+\cos v}{\cos v}$

⇒ $\dfrac{\cos v}{1+\cos v} dv =dx$

Integrating, $\int \dfrac{\cos v}{1+\cos v} dv =\int dx+C$

⇒ $\int \dfrac{1+\cos v -1}{1+\cos v} dv =x+C$

⇒ $\int dv – \int \dfrac{1}{1+\cos v} dv =x+C$

⇒ $\int dv – \int \dfrac{1}{2\cos^2 \frac{v}{2}} dv =x+C$, by the formula 1+cos2x = 2cos^{2}x.

⇒ $v – \int \dfrac{1}{2}\sec^2 \frac{v}{2} dv =x+C$

⇒ $v – \tan \dfrac{v}{2} =x+C$

⇒ $x+y – \tan \dfrac{x+y}{2} =x+C$ as v=x+y.

⇒ $y = \tan \dfrac{x+y}{2}+C$.

So the solution of dy/dx =sec(x+y) is equal to y =tan (x+y)/2 +C where C is an integration constant.

**More Differential Equations:**

## FAQs

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

**Answer:** The solution of dy/dx=sec(x+y) is given by y =tan (x+y)/2 +C where C is a constant.