Integrating Factor

The Integrating Factor method is a powerful technique for solving first-order linear differential equations. It involves multiplying the differential equation by a specially chosen function, the integrating factor, to simplify and solve the equation. This method is also applicable to certain second-order linear differential equations, making it a versatile tool in mathematical problem-solving.

1.0Integrating Factor

Differential equations are fundamental in numerous fields such as physics, engineering, and economics. One of the powerful techniques for solving linear differential equations is the Integrating Factor Method. This method is particularly useful for first-order linear differential equations, but it can also be applied to certain second-order equations. In this blog, we'll delve into the details of this method, explore its application, and provide examples and practice problems to enhance your comprehension.

2.0What is the Integrating Factor Method?

The Integrating Factor Method (I.F) is a technique used to solve first-order linear differential equations of the form:

$\frac{dy}{dx}+P(x)y=Q(x)$

The main idea is to multiply each sides of the differential equation by the integrating factor, which is a specially chosen function, to simplify the equation into a form that is easy to integrate.

3.0Solving the Differential Equation

The most general form of a linear differential equation of first order is $\frac{dy}{dx}+Py=Q$ , where P & Q are functions of x or constant.

To solve such an equation, multiply both sides by $e^{\int Pdx}$ .

So that we get 

$e^{\int Pdx}\left[\frac{dy}{dx}+Py\right]=Q{e}^{\int Pdx}$ ...(i)

$\frac{d}{dx}\left(e^{\int Pdx}\cdot y\right)=Q{e}^{\int Pdx}$ ...(ii)

On integrating equation (ii), we get 

$y{e}^{\int Pdx}=\int Q{e}^{\int Pdx}dx+c$

Or  $y\times I.F=\int (I.F\times Q)dx+c$

Where I.F is $e^{\int Pdx}$

This is the required general solution.

4.0Some Important Points of Integrating Factor

(i) The factor $e^{\int Pdx}$ on multiplying by which the left hand side of the differential equation becomes the differential coefficient of some function of x & y, is called the integrating factor of the differential equation popularly abbreviated as  I.F.

(ii) Sometimes a given differential equation becomes linear if we take y as the independent variable and x as the dependent variable. e.g. the equation; $(x+y+1)\frac{dy}{dx}=y^2+3$ can be written as

$\left(y^2+3\right)\frac{dx}{dy}=x+y+1$, which is a linear differential equation.

5.0Integrating Factor Solved Example 

Example 1: Solve $\left(1+{\mathrm{y}}^2\right)+\left(\mathrm{x}-{\mathrm{e}}^{{\tan }^{-1}\mathrm{y}}\right)\frac{dy}{dx}=0$.

Solution:

Differential equation can be rewritten as ${\left(1+{\mathrm{y}}^2\right)}^{\frac{dx}{dy}}+\mathrm{x}=e^{{\tan }^{-1}y}$

$or\frac{dx}{dy}+\frac{1}{1+y^2}\cdot x=\frac{e^{{\tan }^{-1}y}}{1+y^2}$ ...(i)

I. F = $e^{\int \frac{1}{1+y^2}dy}=e^{{\tan }^{-1}y}$

So, solution is $x{e}^{{\tan }^{-1}y}=\int \frac{e^{{\tan }^{-1}y}{e}^{{\tan }^{-1}y}}{1+y^2}dy$

Let $e^{{\tan }^{-1}y}=t\Rightarrow \frac{e^{{\tan }^{-1}y}}{1+y^2}dy=dt$

$x{e}^{{\tan }^{-1}y}=\int tdt$

or $x^{e^{\tan -1}y}=\frac{t^2}{2}+\frac{c}{2}\Rightarrow 2x^{e^{{\tan }^{-1}y}}=e^{2{\tan }^{-1}y}+c$.

Example 2: The solution of differential equation ${\left({\mathrm{x}}^2-1\right)}^{\frac{dy}{dx}}+2xy=\frac{1}{x^2-1}$ is -

(A) $y\left(x^2-1\right)=\frac{1}{2}\log \left|\frac{x-1}{x+1}\right|+C$

(B) $y\left(x^2+1\right)=\frac{1}{2}\log \left|\frac{x-1}{x+1}\right|-C$

(C) $y\left(x^2-1\right)=\frac{5}{2}\log \left|\frac{x-1}{x+1}\right|+C$

(D) none of these

Solution:

Ans.(A)

The given differential equation is

$\left(x^2-1\right)\frac{dy}{dx}+2xy=\frac{1}{x^2-1}\Rightarrow \frac{dy}{dx}+\frac{2x}{x^2-1}y=\frac{1}{{\left(x^2-1\right)}^2}$    ...(i)

This is linear differential equation of the form

$\frac{dy}{dx}+Py=Q$ ; where $P=\frac{2x}{x^2-1}$ and $Q=\mathrm{Q}=\frac{1}{{\left(x^2-1\right)}^2}$

∴ I.F. = $e^{\int Pdx}=e^{\int 2x/\left(x^2-1\right)dx}=e^{\log \left(x^2-1\right)}=\left(x^2-1\right)$

multiplying both sides of (i) by I.F. = (x² – 1), we get

$\left(x^2-1\right)\frac{dy}{dx}+2xy=\frac{1}{x^2-1}$

integrating both sides we get

$y\left(x^2-1\right)=\int \frac{1}{x^2-1}dx+C$ [Using: y (I.F.) = $\int Q\cdot (I.F.)dx+C$]

$\Rightarrow \mathrm{y}\left({\mathrm{x}}^2-1\right)=\frac{1}{2}\log \left|\frac{x-1}{x+1}\right|+C$.

Example 3: (x² + y)dx – x dy = 0

Solution:

⇒ (x2 + y)dx = xdy

⇒ $\left(x^2+y\right)=\frac{xdy}{dx}$

$\Rightarrow \frac{xdy}{dx}=x^2+y$

$\Rightarrow \frac{dy}{dx}=\frac{x^2+y}{x}\text{ (Divide by }x\text{ ) }$

⇒ $\frac{dy}{dx}-\frac{y}{x}=x$ [Taking term of y & x in L.H.S]

So now $P(x)=\frac{-1}{\mathrm{x}}$ , Q(x) = x

$\Rightarrow \frac{dy}{dx}+P(x)y=Q(x)$

Comparing P(x) & Q(x) from the equation.

If = $e^{\int Pdx}=e^{\int -\frac{1}{x}dx}=e^{-\log x}=e^{\log x^{-1}}=\frac{1}{x}$

Putting the value in the equation.

$y(I.F)=\int Q(x)\times (\text{ I.F })dx+c$.

$y\left(\frac{1}{x}\right)=\int x\cdot \frac{1}{x}dx+c.$

$y\left(\frac{1}{x}\right)=\int dx+c$

$y\left(\frac{1}{x}\right)=x+cy=x^2+cx.$

This is the solution to the differential equation.

Example 4: ${\cos }^{2}x\frac{dy}{dx}+y=\tan x\left(0\le x<\frac{\pi }{2}\right)$

Solution:

$\frac{dy}{dx}+\frac{1}{{\cos }^{2}x}y=\frac{\tan x}{{\cos }^{2}x}$ [divide by cos²x]

$\frac{\mathrm{dy}}{\mathrm{dx}}$+ sec²xy = tanx · sec²x.

So by comparing with equation

$\frac{dy}{dx}+P(x)y=Q(x)$

P(x) = sec²x Q(x) = tanx sec²x.

I.F = ${\mathrm{e}}^{\int \mathrm{Pdx}}={\mathrm{e}}^{\int {\sec }^{2}xdx}={\mathrm{e}}^{\tan x}$

Putting in the equation

y(I.F) = $\int Q(x)\times (\text{ I.F })dx+c$

${\mathrm{ye}}^{\tan x}=\int \tan x\cdot {\sec }^{2}x\cdot e^{\tan x}dx+c$

$I=\int \tan x\cdot {\sec }^{2}x\cdot e^{\tan x}dx.$

Let tan x = t ⇒ sec²xdx = dt.

$I=\int t\cdot e^{t}dt=(t-1)e^t$

$I=(\tan x-1)e^{\tan x}$

ye^tanx = (tanx – 1)e^tanx + c.

y = tanx – 1 + ce^tanx

Example 5: (1 + x²)dy + 2xydx = cotx dx (x ≠ 0)

Solution:

(1 + x²)dy = (cotx – 2xy)dx

${\left(1+x^2\right)}^{\frac{dy}{dx}}=\cot x-2xy$

${\left(1+x^2\right)}^{\frac{dy}{dx}}+2xy=\cot x$

$\frac{dy}{dx}+\frac{2x}{\left(1+x^2\right)}y=\frac{\cot x}{\left(1+x^2\right)}$

Comparing with equation $\frac{dy}{dx}+P(x)y=Q(x)$

$P(x)=\frac{2x}{\left(1+x^2\right)},\mathrm{Q}(\mathrm{x})=\frac{\cot x}{\left(1+{\mathrm{x}}^2\right)}$

I.F. = $e^{\int Pdx}=e\int \frac{2x}{\left(1+x^2\right)}dx=e^{\log \left(1+x^2\right)}=1+x^2$

Putting in equation

y(I.F) = ∫(Q × I.F)dx + c

y(1 + x²) = $\int \frac{\cot x}{\left(1+x^2\right)}\times \left(1+x^2\right)dx+c$

y(1 + x²) = ∫cotx dx + c

y(1 + x²) = log |sin x| + c

Example 6: The integrating factor of the differential equation : (1 – y²) + yx = ay(–1 < y < 1) is

Solution:

(1 – y²) + yx = ay.

$\frac{dx}{dy}+\frac{y}{\left(1-y^2\right)}x=\frac{ay}{\left(1-y^2\right)}$

Comparing with equation $\frac{dx}{dy}+P(y)x=Q(y)$

I.F = $e^{\int p(y)dy}$

$P(y)=\frac{y}{\left(1-y^2\right)},Q(y)=\frac{y}{\left(1-y^2\right)}$

$=e^{\int \frac{y}{1-y^2}\mathrm{dy}}={\mathrm{e}}^{-\frac{1}{2}\log \left(1-y^2\right)}=e^{\log \left(-\frac{1}{\sqrt{1-y^2}}\right)}$

$=-\frac{1}{\sqrt{1-y^2}}$

So, the integrating factor of the equation is $-\frac{1}{\sqrt{1-y^2}}.$

6.0Integrating Factor Practice problems 

1. Solve the differential equation $\frac{dy}{dx}+3y=6$ .
2. Find the integrating factor and solve $\frac{dy}{dx}-y=x$
3. Apply the integrating factor method to solve $\frac{dy}{dx}+y\cot (x)=\sin (x)$.
4. Solve the differential equation $\frac{dy}{dx}+y=e^x$.
5. Solve the differential equation $\frac{dy}{dx}-3y=6$.
6. Solve the differential equation $x\frac{dy}{dx}+y=x^2$.
7. Solve the differential equation $\frac{dy}{dx}+2xy=x$

7.0Solved Questions on Integrating Factor 

1. What is an Integrating Factor?

Ans: An integrating factor is a function used to simplify the solution of a linear differential equation. For a first-order linear differential equation of the form $\frac{dy}{dx}+P(x)y=Q(x)$, the integrating factor I.F is $e^{\int P(x)dx}$ .

2. How do you find the Integrating Factor for a differential equation?

Ans: To find the integrating factor I.F for the first-order linear differential equation $\frac{dy}{dx}+P(x)y=Q(x)$ , compute: $\text{ I.F }=e^{\int P(x)dx}$

3. How do you use the Integrating Factor to solve a differential equation?

Ans: To solve $(\frac{dy}{dx}+P(x)y=Q(x)$ using the integrating factor:

• Compute the integrating factor: $\text{ I. F }=e^{\int P(x)dx}$ .
• Multiply the entire differential equation by I. F.
• Recognize the left-hand side as the derivative of I. F: $\frac{d}{dx}[(I.F)y]=(I.F)Q(x)$ .
• Integrate both sides: $(I.F)\cdot y=\int (I.F)\cdot Q(x)dx$ .
• Solve for y.