What kind of ODE is this and how to solve

Mary Keefe

Mary Keefe

Answered question

2022-01-22

What kind of ODE is this and how to solve it?
(2x4y+6)dx+(x+4y3)dy=0
I converted it to this form
dydx=2x4y+6x+4y3 in the hope I can use z=yx substitution for homogeneous ODEs but because of constants 6 and 3 I can't.

Answer & Explanation

Jenny Sheppard

Jenny Sheppard

Beginner2022-01-22Added 35 answers

Do first a change of variables:
u=x+1 v=y1
Then du=dx and dv=dy. Now
2u4v=2x+24y+4=2x4y+6
and u+4v=x+1+4y4=x+4y3
So in the new variables you are solving
(2u4v)du+(u+4v)dv=0
or (away from the line u+4v=0)
2u4vu+4vdu+dv=0
Then you can do the substitution: write v=uz you get dv=zdy+udz:
(24z1+4z+z)du+udz=0
Robert Pina

Robert Pina

Beginner2022-01-23Added 42 answers

This is an ODE that can be solved in the following way.
Write (2x4y+6)dx+(x+4y3)dy=0
as 2x+4y6x+4y3=dydx
What we need now is to write the RHS as
ax+bycx+dy
so that it becomes homogeneous.
So we need to find k and h such that
2(x1+h)+4(y1+k)6=2x1+4y1
(x1+h)+4(y1+k)3=x1+4y1
This is, we need that
2h+4k6=0
h+4k3=0
This gives, h=1
k=1
So we have
2x1+4y1x1+4y1=dy1dx1
But this ODE is homogeneous (as we wanted), so we can put
y1x1=v
and get 2+4v1+4v=vx1+v
This results in a separable ODE,
dx1x1=4v+14v23v+2dv
I guess you can take it from here.
NOTE: In general, the equation
ax+by+cdx+ey+f=dydx
can be made homogeneous by solving
ah+bk+c=0
dh+ek+f=0
and substituting X=x+h,Y=y+k.
In turn, the homogeneous ODE can be made separable by putting YX=v.

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