Solving this differential equation

[chwala]

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]##

 

[chwala]

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 $B=-0.5$

 

[Fred Wright]

You have
$$
\int \frac{\frac{4}{3}-v}{v^2-\frac{8}{3}v-1}dv
$$
what are the roots of $v^2-\frac{8}{3}v-1$?

 

[Mark44]

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.

$-\frac {dx}{x}=\frac {4-3v}{8v-3v^2+3}$
$\frac {dx}{x}=\frac {4-3v}{3v^2-8v-3}$=[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.

 

[Chestermiller]

Isn't your original relationship an exact differential: $$3xdx+4d(xy)-3ydy=0$$

 

[chwala]

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

 

[chwala]

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...

 

[chwala]

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?...

 

[chwala]

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...

 

[Mark44]

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.

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

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.

 

[chwala]

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...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...

 

[chwala]

which means my post $2$ was just correct after all.

 

[Mark44]

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.

 

[chwala]

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...

 

[chwala]

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.

 

[Mark44]

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...

$lnx$= 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.

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

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...

 

[Mark44]

$c=\frac {3x^2}{2}+4xy-\frac {3x^2}{2}$ where $c$ is an arbitrary constant.

Exactly what I got.
Edit: That last term should be $-\frac 3 2 y^2$.

 

[chwala]

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...

 

[chwala]

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

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

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

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

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

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

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

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

→$(3y+x)^3=y-3x$

 

[chwala]

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...

 

[Mark44]

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.

 

[chwala]

wait... but even in the exact approach, the first and third terms cancel out, leaving us with
$c=4xy$

 

[Mark44]

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.

 

[chwala]

i amended mine too, ...saw the error ...lol

 

[chwala]

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.

 

[chwala]

my other attempt on this...

$ln x= -1.5[ ln (3v+1)+0.5 ln (v-3)]+ln c$

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

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

$\frac{x(3y+x)^{1.5}(-3x+y)^{0.5}}{x^2}=c$

$\frac{(3y+x) (3y+x)^{0.5}(-3x+y)^{0.5}}{x}=c$
$\frac{(3y+x) (-9xy+3y^2-3x^2+xy)^{0.5}}{x}=c$
$\frac{(3y+x) (3y^2-8xy-3x^2)^{0.5}}{x}=c$

 

[Mark44]

now, i have re done the work for my post $19$ as follows,
$ln x= -1.5[ ln (3v+1)+0.5 ln (v-3)]$

You can't omit the constant of integration! And you have a pair of brackets that shouldn't be there.
This should be $\ln x= -1.5 \ln (3v+1)+0.5 \ln (v-3) + C$

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

Sign error in second term above.
Corrected the sign error and added in the constant.
$\ln x +1.5 \ln (3v+1) - 0.5 \ln (v-3) = C$

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

Continue from my work above to
$\ln[\frac{x(3v + 1)^{3/2}}{(v - 3)^{1/2}} ] = C$
The next step will be
$\frac{x(3v + 1)^{3/2}}{(v - 3)^{1/2}} = e^C$
$e^C$ is just another constant, call it A.
So you have $\frac{x(3v + 1)^{3/2}}{(v - 3)^{1/2}} = A$
Replace v by y/x. Do not square both sides -- the factors in x will cancel, and things turn out as they should.

$\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.

No, there are a number of errors, as noted above. Also, you should end up with an equation; namely,
$\frac {3x^2}{2}+4xy-\frac {3y^2}{2} = C$

 

[chwala]

Mark after integration, aren't we supposed to have
$ln x= -1.5 ln (3v+1)-0.5 ln (v-3)+ln C$?..I spent time pondering over that.
giving me,
$ln x+1.5 ln (3v+1)+0.5 ln (v-3)=ln C$
and are we not supposed to consider the constant of integration as an $ln$ function i.e $ln C$and not just a constant as you have indicated? or it does not matter?

 

[Mark44]

Mark after integration, aren't we supposed to have
$ln x= -1.5 ln (3v+1)-0.5 ln (v-3)+ln C$?..I spent time pondering over that.
giving me,
$ln x+1.5 ln (3v+1)+0.5 ln (v-3)=ln C$

Yes, this looks fine, and is nearly the same as what you have in post #14.
However, the 2nd line in post #14 has an error.

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

It should be $\ln x = \ln[(3v + 1)^{-1.5}(v - 3)^{-0.5}] + \ln C$
$\Rightarrow \ln x = \ln \left( \frac{(3v + 1)^{-1.5}}{(v-3)^{0.5}}\right) + \ln C$
I have 3 pages of work, both sides, on this problem, and worked two different ways, so there are lots of opportunities to get a sign wrong, plus if you have a sign error in your work somewhere, and I continue from what you showed, that fouls things up as well.

and are we not supposed to consider the constant of integration as an $ln$ function i.e $ln C$and not just a constant as you have indicated? or it does not matter?

 

[chwala]

let me check this again tomorrow, this problem has really mixed me up...