How to solve Homogeneous Differential Equations

By Prashant Kumar|Updated : August 15th, 2018

A function f (x, y) is called homogeneous function of degree n if

For Example:

Here

So this is homogeneous function of degree 4.

To solve this type of question we put

 or

depending on the question.

After putting this we use the same method we use earlier in the question. And we get differential equation in z and x(or y) and then we solve that differential equation and after solution we put back value of z.

 

Example: Solve

Solution:

So this is a homogeneous equation. Here we put

                                          y = vx

                              dy = v dx + x dv

So the differential equation is

Integrate both sides

Now put back value of v = y/x

DIFFERENTIAL EQUATION REDUCIBLE TO HOMOGENEOUS FORMS

We see in the 3rd point about the homogeneous from of equation. Here we see another equation that can be reducible to homogenous from.

Now we can reduce to homogeneous form. We have seen this type of equation earlier but there ‘aB = bA’ So these a little difference between both.

So to solve this type of differential equation we put

                                    x =X+h  and y =Y+k
here h and k are constants and we chose these constants such that

So the new equation is

                       

So here we chose h and k such that they satisfy the equations

                        Ah+bk+c = 0 and Ah+Bk+C = 0

So the value of h and k are

and we can find these value only if the denominator is NOT zero, hence the above condition.

Now if we find these value h and k then the equation become:

Now here we can use the Homogeneous form solution by putting

                        Y = VX or X = VY

After this, the rest of the solution remains the same as discussed in the previous section.

 

Example:

Solution:

Here we can see that ½ is not equal to 2/1. So We use the above method to solve this question.

Let’s put

x = X+h and y = Y+k, and

 

So the equation is

So here h and k be such that  h+2k-5 = 0 and  2h+k-4 = 0

Put k from 2nd to 1st

h+2(4-2h)-5 =0

h+8-4h-5=0

Which gives, h=1 and k=2

So the equation is

Now put Y=VX   so   dY = VdX + XdV

Equation is

So now we integrate both sides.

                         

Put back V = Y/X

So

Here C = c2

Now put

X = x-1 and Y = y-2

So this is our final answer. Here we use substitutions twice to solve this.

Stay tuned for more.

Click on the links below to read more about Differential Equations:

How to Solve Differential Equations by Variable Separable Method

How to Solve Linear Differential Equations

Order, Degree & Solutions of Differential Equations

Click on the links below to access the list:

Daily Quizzes

JEE Weekly Plans

JEE Free Full Length Mock Tests

Download Gradeup, the best IIT JEE Preparation App
Attempt subject wise questions in our new practice section

All the best!!

Team Gradeup

Posted by:

Prashant KumarPrashant KumarMember since Jun 2018
Life is a journey, not a destination. Enjoy it!
Share this article   |

Comments

write a comment
Load Previous Comments
Uriti Manish

Uriti ManishJan 20, 2018

Awesome sir
Uriti Manish

Uriti ManishJan 20, 2018

Very useful
Raees Gaurav Singh
Great sir 👏👏👏👏👏🙏🙏
Subhasish Rabikanta Bal
THANKS  SIR........MAY U LIVE LONG.....SIR
Jasibul Sk

Jasibul SkAug 16, 2018

Thanks .make more video for held us.
jayesh shah

jayesh shahAug 16, 2018

Please request more video of topic integration
KS VASUDHA

KS VASUDHAAug 17, 2018

Y/x ko v Maan ke karna padega
Aryan Anand

Aryan AnandJun 24, 2019

Solve the above question

Follow us for latest updates