How to Find Integrating Factor [with Solved Examples]

The general form of linear differential equation of first order is of the form: dy/dx + P(x)y = Q(x) where P(x) and Q(x) are functions of x. Using integrating factor method, we solve this equation. In this article, let us learn how to solve a first order linear differential equation with examples.

Integrating Factor of dy/dx + P(x)y = Q(x)

The integrating factor of dy/dx + P(x)y = Q(x) is equal to $e^{\int P(x)~dx}$.

How to Solve dy/dx + P(x)y = Q(x)

$\boxed{\textbf{Question:}}$ Solve $\dfrac{dy}{dx}$+P(x)y=Q(x)

Answer:

As the integrating factor of dy/dx + P(x)y = Q(x) is $e^{\int P(x)~dx}$, we multiply both sides of the equation by it. Therefore, we have

⇒ $e^{\int P(x)~dx}$ $\left( \dfrac{dy}{dx}+P(x)y \right)$ = Q(x) $e^{\int P(x)~dx}$

⇒ $\dfrac{d}{dx} \left(y e^{\int P(x)~dx} \right)$ = Q(x) $e^{\int P(x)~dx}$

Integrating,

$\displaystyle \int d \left(y e^{\int P(x)~dx} \right)$ $= \displaystyle \int Q(x) e^{\int P(x)~dx} dx$ + C

⇒ $y e^{\int P(x)~dx}$ $= \displaystyle \int Q(x) e^{\int P(x)~dx} dx$ + C

where C is an arbitrary constant of integration. This is the solution of the linear differential equation dy/dx + P(x)y = Q(x).

Note: 1. The integrating factor (IF) of dy/dx + P(x)y = Q(x) is equal to e∫P(x)dx . 2. The solution of the linear differential equation dy/dx + P(x)y = Q(x) of first order is given as follows: y × IF = ∫{IF × Q(x)}dx + C.

Solved Problems

Question 1: Solve dy/dx – y = ex.

Answer:

Comparing dy/dx – y = e^x with dy/dx + P(x)y = Q(x), we have that:

P(x) = -1 and Q(x) = e^x.

So Integrating Factor (IF) = e^∫P(x)dx = e^{∫-1 dx} = e^-x.

Therefore, the general solution is given by

y × IF = ∫{IF × Q(x)}dx + C

⇒ ye^-x = ∫{e^-x × e^x}dx + C

⇒ ye^-x = ∫dx + C

⇒ ye^-x = x + C

⇒ y = xe^x + Ce^x where C is an arbitrary constant of integrations.

Question 2: Solve dy/dx – y/x = xex.

Answer:

Comparing dy/dx – y/x = xe^x with dy/dx + P(x)y = Q(x), we have that:

P(x) = -1/x and Q(x) = xe^x.

So Integrating Factor (IF) = e^∫P(x)dx = e^{∫-1/x dx} = e^{-ln x} = $e^{\ln x^{-1}}$ = x^-1 = 1/x.

Therefore, the general solution is given by

y × IF = ∫{IF × Q(x)}dx + C

⇒ y/x = ∫{xe^x × 1/x}dx + C

⇒ ye^-x = ∫e^x dx + C

⇒ ye^-x = e^x + C where C is an arbitrary constant.

Practice Problems:

Solve the following linear differential equations of first order:

dy/dx + y = x

Engineering Mathematics II HomePage

FAQs