Solving this differential equation

  • Thread starter chwala
  • Start date
  • #1
chwala
Gold Member
796
64

Homework Statement:

##(3x+4y)dx + (4x-3y)dy=0##

Relevant Equations:

##\frac{dy}{dx}##= ##v##+##x\frac {dv}{dx}##
##-\frac {dy}{dx}=\frac {3+4v}{4-3v}##
##\frac {3+4v}{4-3v}=-v-x\frac {dv}{dx}##
##-\frac {dx}{x}=\frac {4-3v}{8v-3v^2+3}##
##\frac {dx}{x}=\frac {4-3v}{3v^2-8v-3}##=[A/3v+1]+[B/v-3]##
 
Last edited:

Answers and Replies

  • #2
chwala
Gold Member
796
64
Homework Statement:: ##(3x+4y)dx + (4x-3y)dy=0##
Relevant Equations:: ##\frac{dy}{dx}##= ##v##+##x\frac {dv}{dx}##

##-\frac {dy}{dx}=\frac {3+4v}{4-3v}##
##\frac {3+4v}{4-3v}=-v-x\frac {dv}{dx}##
##-\frac {dx}{x}=\frac {4-3v}{8v-3v^2+3}##
##\frac {dx}{x}=\frac {4-3v}{3v^2-8v-3}##=[A/3v+1]+[B/v-3]##
ok, are my steps correct? then i can proceed. Actually i have a solution that is different from textbook, that's why i need a second perspective from you...i have ##A=-1.5## and ##B=-0.5##
 
  • #3
274
135
You have
$$
\int \frac{\frac{4}{3}-v}{v^2-\frac{8}{3}v-1}dv
$$
what are the roots of ##v^2-\frac{8}{3}v-1##?
 
  • #4
34,288
5,925
Homework Statement:: ##(3x+4y)dx + (4x-3y)dy=0##
Relevant Equations:: ##\frac{dy}{dx}##= ##v##+##x\frac {dv}{dx}##

##-\frac {dy}{dx}=\frac {3+4v}{4-3v}##
##\frac {3+4v}{4-3v}=-v-x\frac {dv}{dx}##
How did you get the line above? I would guess that you are doing something related to your Relevant Equations.
If so, the assumption there is that y = vx, so dy = vdx + xdv, and hence ##\frac{dy}{dx} = v + x\frac{dv}{dx}##.
Please show what you are substituting for v, and include the work that leads to the line below.
chwala said:
##-\frac {dx}{x}=\frac {4-3v}{8v-3v^2+3}##
##\frac {dx}{x}=\frac {4-3v}{3v^2-8v-3}##=[A/3v+1]+[B/v-3]##
Also, in the line above, you need more parentheses.
A/3v + 1 should be written as A/(3v + 1), and B/v-3 should be written as B/(v-3). The brackets you included are doing nothing to fix the problem - they need to be around the denominators, not the whole fraction.
 
  • Like
Likes DaveE and berkeman
  • #5
20,589
4,436
Isn't your original relationship an exact differential: $$3xdx+4d(xy)-3ydy=0$$
 
  • Like
Likes chwala
  • #6
chwala
Gold Member
796
64
Thanks guys...let me respond later I finish with some errands, I will show my steps as asked...I didn't check to establish if the equation was exact, ....I will check that also. Regards
 
  • #7
chwala
Gold Member
796
64
How did you get the line above? I would guess that you are doing something related to your Relevant Equations.
If so, the assumption there is that y = vx, so dy = vdx + xdv, and hence ##\frac{dy}{dx} = v + x\frac{dv}{dx}##.
Please show what you are substituting for v, and include the work that leads to the line below.
Also, in the line above, you need more parentheses.
A/3v + 1 should be written as A/(3v + 1), and B/v-3 should be written as B/(v-3). The brackets you included are doing nothing to fix the problem - they need to be around the denominators, not the whole fraction.
I will fix that too, in my subsequent working...
 
  • #8
chwala
Gold Member
796
64
How did you get the line above? I would guess that you are doing something related to your Relevant Equations.
If so, the assumption there is that y = vx, so dy = vdx + xdv, and hence ##\frac{dy}{dx} = v + x\frac{dv}{dx}##.
Please show what you are substituting for v, and include the work that leads to the line below.
Also, in the line above, you need more parentheses.
A/3v + 1 should be written as A/(3v + 1), and B/v-3 should be written as B/(v-3). The brackets you included are doing nothing to fix the problem - they need to be around the denominators, not the whole fraction.
ok this was my approach,

##(3x +4y)dx +(4x-3y)dy=0##

##(3x +4y)dx =-(4x-3y)dy##

##\frac {3x+4y}{4x-3y}=-\frac {dy}{dx}##

Using, ##y=vx##

→##\frac {3x+4vx}{4x-3vx}=-\frac {dy}{dx}##

→##\frac {3+4v}{4-3v}=-\frac {dy}{dx}##

→##\frac {3+4v}{4-3v}=-v-x\frac {dv}{dx}##

→##-\frac {dx}{x}=\frac {4-3v}{8v-3v^2+3}##

→##\frac {dx}{x}=-\frac {4-3v}{8v-3v^2+3}##

→##\frac {dx}{x}=\frac {4-3v}{-8v+3v^2-3}##

does this negative also affect the numerator or is my working just fine?....
 
  • #9
chwala
Gold Member
796
64
Isn't your original relationship an exact differential: $$3xdx+4d(xy)-3ydy=0$$
i will check it out, i have re posted my earlier approach though...
 
  • #10
34,288
5,925
i have A=−1.5 and B=−0.5
I get the same values.

ok this was my approach,
##(3x +4y)dx +(4x-3y)dy=0##
##(3x +4y)dx =-(4x-3y)dy##
##\frac {3x+4y}{4x-3y}=-\frac {dy}{dx}##
Using, ##y=vx##
→##\frac {3x+4vx}{4x-3vx}=-\frac {dy}{dx}##
→##\frac {3+4v}{4-3v}=-\frac {dy}{dx}##
→##\frac {3+4v}{4-3v}=-v-x\frac {dv}{dx}##
→##-\frac {dx}{x}=\frac {4-3v}{8v-3v^2+3}##
Above, you're missing dv from the right hand side.
chwala said:
→##\frac {dx}{x}=-\frac {4-3v}{8v-3v^2+3}##
→##\frac {dx}{x}=\frac {4-3v}{-8v+3v^2-3}##

does this negative also affect the numerator or is my working just fine?....
No, the negative affects only the numerator or only the denominator, but not both. If it were to effect both, that would be equivalent to multiplying by ##\frac {−1}{−1}##, which is different from having a factor of -1.
In your last line, you should add the dv factor, and write the denominator like this:
##\frac {dx}{x}=\frac {4-3v}{3v^2 -8v -3}dv##
Now rewrite the right side using your partial fraction decomposition, like so:
##\frac {dx}{x}=\frac {−3/2}{3v+1}dv+ \frac{−1/2}{v−3}dv##
Integrate both sides to get x in terms of v. Undo the substitution to get y back to end up with an equation that involves x and y.

@Chestermiller's suggestion was probably a lot simpler than what you're doing, but you've done most of the work already.
 
  • Like
Likes chwala
  • #11
chwala
Gold Member
796
64
I get the same values.

Above, you're missing dv from the right hand side.
No, the negative affects only the numerator or only the denominator, but not both. If it were to effect both, that would be equivalent to multiplying by ##\frac {−1}{−1}##, which is different from having a factor of -1.
In your last line, you should add the dv factor, and write the denominator like this:
##\frac {dx}{x}=\frac {4-3v}{3v^2 -8v -3}dv##
Now rewrite the right side using your partial fraction decomposition, like so:
##\frac {dx}{x}=\frac {−3/2}{3v+1}dv+ \frac{−1/2}{v−3}dv##
Integrate both sides to get x in terms of v. Undo the substitution to get y back to end up with an equation that involves x and y.

@Chestermiller's suggestion was probably a lot simpler than what you're doing, but you've done most of the work already.
aaaaaaaaaargh, i really hate making those kind of mistakes:frown:....of leaving out the ##dv##. bingo mark, let me check it out...i will also use Chestermiller's approach and see what comes out of it...
 
Last edited:
  • #12
chwala
Gold Member
796
64
which means my post ##2## was just correct after all.
 
  • #13
34,288
5,925
which means my post ##2## was just correct after all.
Yes, except for the missing dv. My first quote in the previous thread came from your post #2.
 
  • Like
Likes chwala
  • #14
chwala
Gold Member
796
64
ok just to finish on this.....
on integration,i have
##ln x= -1.5 ln(3v+1)-0.5 ln (v-3)##

##lnx##= ln##\frac {(3v+1)^{-1.5}}{(v-3)^{-0.5}}##

##lnx##= ln##\frac {(v-3)^{0.5}}{(3v+1)^{1.5}}##

→##x##= ##\frac {(v-3)^{0.5}}{(3v+1)^{1.5}}##

→##x.\frac {(3x+y)^{1.5}}{(y)^{1.5}}##=##\frac {(x-3y)^{0.5}}{(y)^{0.5}}##

→##x^2.\frac {(3x+y)^3}{(y)^{1.5}}##=##\frac {x-3y}{y}##

looks a bit complicated...
on cross multiplication, we have two scenarios, either,
##y^3(x-3y)=0##
or
##x^2y(3x+y)^3=0##
tell me am doing something wrong...
 
Last edited:
  • #15
chwala
Gold Member
796
64
using the exact approach,

##M=3x+4y## and ##N=4x-3y## where ##Mdx+Ndy=0##

##\frac{∂M}{∂y}=4## and ##\frac {∂N}{∂x}=4## hence exact, therefore,

##U(x,y)= ∫(3x+4y)dx +∫(4x-3y)dy##

##c=\frac {3x^2}{2}+4xy-\frac {3y^2}{2}## where ##c## is an arbitrary constant.
 
Last edited:
  • #16
34,288
5,925
ok just to finish on this.....
on integration,i have
##ln x= -1.5 ln(3v+1)-0.5 ln (v-3)##
Plus the constant of integration...
chwala said:
##lnx##= ln##\frac {(3v+1)^{-1.5}}{(v-3)^{-0.5}}##

##lnx##= ln##\frac {(v-3)^{0.5}}{(3v+1)^{1.5}}##

→##x##= ##\frac {(v-3)^{0.5}}{(3v+1)^{1.5}}##

→##x.\frac {(3x+y)^{1.5}}{(y)^{1.5}}##=##\frac {(x-3y)^{0.5}}{(y)^{0.5}}##
I don't know about the step above. The substitution was y = vx, or v = y/x, so replace v with y/x. I can't tell if you did that.
chwala said:
→##x^2.\frac {(3x+y)^3}{(y)^{1.5}}##=##\frac {x-3y}{y}##

looks a bit complicated...
on cross multiplication, we have two scenarios, either,
##y^3(x-3y)=0##
or
##x^2y(3x+y)^3=0##
tell me am doing something wrong...
 
  • #17
34,288
5,925
##c=\frac {3x^2}{2}+4xy-\frac {3x^2}{2}## where ##c## is an arbitrary constant.
Exactly what I got.
Edit: That last term should be ##-\frac 3 2 y^2##.
 
Last edited:
  • #18
chwala
Gold Member
796
64
Plus the constant of integration...
I don't know about the step above. The substitution was y = vx, or v = y/x, so replace v with y/x. I can't tell if you did that.
aaaaargh stupid mistake on my part...i substituted ##x/y## instead of ##y/x## let me re do it...
 
  • #19
chwala
Gold Member
796
64
##ln x= -1.5 ln(3v+1)-0.5 ln (v-3)##

##lnx##= ln##\frac {(3v+1)^{-1.5}}{(v-3)^{-0.5}}##

##lnx##= ln##\frac {(v-3)^{0.5}}{(3v+1)^{1.5}}##

→##x##= ##\frac {(v-3)^{0.5}}{(3v+1)^{1.5}}##

→##x.\frac {(3y+x)^{1.5}}{x^{1.5}}##=##\frac {(y-3x)^{0.5}}{(x)^{0.5}}##

→##x^2.\frac {(3y+x)^3}{x^3}##=##\frac {y-3x}{x}##

→##x^2.\frac {(3y+x)^3}{x^3}##=##\frac {y-3x}{x}##

→##\frac {(3y+x)^3}{x}##=##\frac {y-3x}{x}##

→##(3y+x)^3=y-3x##
 
  • #20
chwala
Gold Member
796
64
Plus the constant of integration...
I don't know about the step above. The substitution was y = vx, or v = y/x, so replace v with y/x. I can't tell if you did that.
yeah i missed out on the constant as its an indefinite integral...noted...
 
  • #21
34,288
5,925
The first line in post #19 still omits the constant of integration. Also, the last line doesn't match your result in post #15. The solution you got in post #15, ##c=\frac {3x^2}{2}+4xy-\frac {3x^2}{2}##, satisfies the differential equation -- the one in post #19 does not.
I think you have a mistake in the work you did originally, but I can't put my finger on where you went wrong.
 
  • #22
chwala
Gold Member
796
64
wait... but even in the exact approach, the first and third terms cancel out, leaving us with
##c=4xy##
 
  • #23
34,288
5,925
My bad. I misread your post #15. What I got was ##4xy + \frac 3 2 x^2 - \frac 3 2 y^2 = C## I have edited my earlier post.
 
  • #24
chwala
Gold Member
796
64
i amended mine too, ....saw the error ...lol
 
  • #25
chwala
Gold Member
796
64
now, i have re done the work for my post ##19## as follows,
##ln x= -1.5[ ln (3v+1)+0.5 ln (v-3)]##

##ln x + 1.5 ln (3v+1) +0.5 ln (v-3)=0##

##ln x + 1.5 ln (3v+1) +0.5 ln (v-3)= ln 1##

##\frac {(3y+x)^3} {x^3}####\frac {y-3x} {x}.x^2=0## on multiplying both sides by ##x^2##

##x(3y+x)(y-3x)=0##

##(3y^2-9xy+xy-3x^2)=0##

##(3y^2-8xy-3x^2)=0##

on dividing each term by ##2## and multiplying by ##-1##, we get,

##\frac {3x^2}{2}+4xy-\frac {3y^2}{2}##

does this make sense compared to the solution found by exact method?...i see an error in my working, i will look at this again...any insights will be appreciated.
 
Last edited:
Top