# 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.

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