The general solution of the differential equation dy/dx+ytanx = secx is equal to ysecx =tanx +C where C is a constant. This equation is an example of a first order linear ordinary differential equation.

## Solve dy/dx+ytanx = secx

**Question:** Solve the differential equation $\dfrac{dy}{dx}$+ytanx = secx.

**Solution:**

The given equation has the form

$\dfrac{dy}{dx}+P(x)=Q(x)$,

a 1st order linear ordinary differential equation (ode). So the given equation is a first order linear ode with

P(x) = tanx, Q(x) =secx.

Its integrating factor I(x) = e^{∫P(x)dx} = e^{∫tanx dx} = e^{log secx} = secx.

So the solution will be

y I(x) = ∫Q(x) I(x) dx +C

⇒ y secx = ∫secx secx dx +C

⇒ y secx = ∫sec^{2}x dx +C

⇒ y secx = tanx dx +C

So the general solution of the ode dy/dx+ytanx = secx is equal to y secx = tanx dx +C where C is an arbitrary integral constant.

**More Problems:** Solve dy/dx =1+x+y+xy

## FAQs

### Q1: What is the solution of dy/dx+ytanx = secx?

**Answer:** The solution of dy/dx+ytanx = secx is given by y secx = tanx dx +C where C denotes a constant.