Linear Differential Equation

A linear differential equation is a differential equation that is defined by a linear equation or polynomial with one or more terms in the unknown function and its derivatives of dependent variable. The standard form of the first order, commonly known as Leibnitz’s linear equation, is

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

where P(x), Q(x) are the functions of variable ‘x’.


What is linear differential equation?

A linear differential equation is a differential equation that is defined by a linear equation or polynomial with one or more terms in the unknown function and its derivatives of dependent variable. A differential equation of type

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

is called as linear differential equation where P and Q are either constants or functions of x only. This type of differentials equation is first order differential equation as order of derivative is one. Some of the examples of linear differential equation are


$1) \frac{dv}{dt}=\tan x -y$

$2) x \log x\frac{dy }{dx}+y= e^x$

The value of dependent variable y is the solution of the linear differential equation. So, let's see how to find solution to a linear first order differential equation. Remember as we go through this process that the aim is to find a solution that is in the form   y=y(x) In order to find solution, we have derived a formula for the general solution as below.

Non-linear Differential Equation

A non-linear differential equation is one in which the unknown function and its derivatives don’t have a straight line when plotted in a graph (the linearity or non-linearity in the arguments of the function are not considered here). There aren’t many ways to solve nonlinear differential equations exactly, and the ones that do exist usually require the equation to have certain symmetries.

General Solution of linear differential equation

In order to solve a linear first order differential equation, the differential equation must in the form shown below.

$$\frac{dy}{dx}+P(x)y= Q(x)~~~~~~~-(1)$$                                  

Where both P(x) and Q(x) are continuous functions. A continuous function is function if  no holes or breaks in function.

Now, assume that there is some function called , R(x) , called an integrating factor. Do not fear about what this function is or where it came from at this point. After finding out the formula for the general solution , we will find out what R(x).

So, now that we have assumed the existence of R(x) , multiply both sides in (1) by R(x). We will get

$$R(x)\frac{dy}{dx}+P(x)R(x)y=R(x) Q(x)~~~~~~-(2)$$      

Now, this is where the magic of R(x) comes into play. We are going to assume that whatever is R(x), it will satisfy the following.

$$P(x)R(x)=R'(x)~~~~~~-(3)$$                                                         

Again do not think about how we can find a R(x) that will satisfy (3). As we will see, provided P(x) is continuous we can find it. So substituting (3) we now come at.

$$R(x)\frac{dy}{dx}+R'(x)y=R(x) Q(x)~~~~~~~-(4)$$              

In the above product rule is followed by the left hand side of  (4) nothing more than that.

$$R(x)\frac{dy}{dx}+R'(x)y=(R(x) y(x))'$$        

So use product rule to replace the left hand side of (4). By substituting this (4) becomes as follows.

$$(R(x)y(x))'=R(x)Q(x)~~~~~~~~~-(5)$$

Now, recall that we have to find  y(x). All we need to do is integrate both sides then use a little algebra to find out the solution. So, integrate both sides of (5) to get.

$$\int{(R(x)y(x))'}dx=\int{R(x)Q(x)}dx$$

$$R(x)y(x)+c=\int{R(x)Q(x)}dx~~~~~~~~~-(6)$$

It is very important to add the constant of integration, c, from the left side of integration. Solution of differential equation will get wrong if correctly not used.

The final step is then some algebra to solve for the solution, y(x)

$$R(x) y(x)= \int {R(x)Q(x)} dx -c $$

$$y(x)= \frac{\int {R(x)Q(x)} dx -c }{R(x)}$$

Now, from a notational standpoint, the constant of integration, c, is an unknown constant and so to make solution easy we will change the minus sign in front of it into the constant and use a plus instead. No change will be in this solution. So with this change we have.

$$y(x)= \frac{\int {R(x)Q(x)} dx +c }{R(x)}~~~~~~-(7)$$

Again, we keep the minus sign of constant of integration, c, we will get the same value of c as per expectation except it will have the opposite sign.

So, to find out the general solution to linear differential equation, need to go back and determine just what this function R(x)  is. Now start from (3).

$$P(x)R(x)=R'(x)$$

Divide both sides by R(x),

$$\frac{R'(x)}{R(x)}=P(x)$$

Now, the left side of this is derivative as per $\frac{d}{dt}(\log t)=\frac{1}{t}$ with t as R(x), so change above left side accordingly,

$$(\log R(x))'=P(x)$$

Integrate both sides of the above to get,

$$\log R(x) +k=\int{P(x)} dx$$

$$\log R(x) =\int{P(x)} dx +k $$

Now again see that the constant of integration from the left-hand side, k, had been moved to the right-hand side and minus sign replace again as per earlier. So, to avoid confusion different letters k and c are used since they will have different values in the solution we finding out.

To move out R(x) of the natural logarithm, exponentiate both sides of the above,

$$R(x)= e^{\int{P(x)} dx +k}$$

$$R(x)= e^k \dot e^{\int{P(x)} dx +k}$$

Since we know that k is an unknown constant.  $e^k$can be taken as unknown constant k . This will give us the following.

$$R(x)=k e^{\int{P(x)} dx }~~~~~~~-(8)$$

So, a formula for the general solution, (7) is found out and a formula for the integrating factor, (8) also figure out.

Substitute  $ R(x)=k e^{\int{P(x)} dx }$    into (7) and rearrange the constants, 

$$y(x)= \frac{\int {R(x)Q(x)} dx +c }{R(x)}$$

$$y(x)= \frac{\int {Q(x) k e^{\int{P(x)} dx } dx +c} }{k e^{\int{P(x)} dx }}$$

$$y(x)= \frac{k\int {Q(x)e^{\int{P(x)} dx } dx +c }}{k e^{\int{P(x)} dx }}$$

$$y(x)= \frac{\int {Q(x)e^{\int{P(x)} dx } dx +c }}{e^{\int{P(x)} dx }}$$

Solution of a linear first order differential equation is then

$$y(x)= \frac{\int {R(x)Q(x)} dx +c }{R(x)}$$

where Integrating factor,

$$R(x)= e^{\int{P(x)} dx }$$

Solution Steps for linear differential equation

The solution steps for a first order linear differential equation is as follows.

1)    Rearrange the differential equation in the Standard form,


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


2)    Find the integrating factor, R(x) , using 

$$R(x)= e^{\int{P(x)} dx }$$

3)    Multiply both sides in the differential equation by R(x) and verify that the left side becomes the product rule  and write it as such.

4)    Integrate both sides, make sure you properly deal with the constant of integration.

5)    Solve for the solution y(x)

Frequently Asked Questions – FAQ

1)     What is linear differential equation?

      A linear differential equation is a differential equation that is defined by a linear polynomial in the unknown function and its derivatives.

2)     What is standard form of first order linear differential equation?

       The standard form of first order linear differential equation is given as

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

        Where P(x) and Q(x) are continuous functions of x

3)     What is integrating factor for linear differential equation?

        The integrating factor for linear differential equation is given by

$$R(x)= e^{\int{P(x)} dx }$$

4)     What is general solution for linear differential equation?

        Solution of a linear first order differential equation is then

$$y(x)= \frac{\int {R(x)Q(x)} dx +c }{R(x)}$$

         where Integrating factor,   

$$R(x)= e^{\int{P(x)} dx }$$

5)    What are the 4 types of differential equations?

      Some types of differential equation are as: 

  •          Ordinary Differential Equations 
  •          Partial Differential Equations. 
  •          Linear Differential Equations.
  •          Nonlinear differential equations.  
  •          Homogeneous Differential Equations.  
  •         Nonhomogeneous Differential Equations.

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