Solution to DiffEq

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  • #1
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Main Question or Discussion Point

I've come across this problem while self studying Ordinary Differential Equations and I really need help. The problem asks me to simply eliminate derivatives, I do not need to separate. The book shows the answer, but not the steps.

problem:
[itex]\frac{dy}{dx}[/itex]= [itex]\frac{y^2}{1-xy}[/itex]

answer:
ln(y)=xy

So far, I'm here...

[itex]\frac{dy}{dx}[/itex]= [itex]\frac{y^2}{1-xy}[/itex]
y' = [itex]\frac{y^2}{1-xy}[/itex]
y' = [itex]\frac{y}{\frac{1}{y}-x}[/itex]
y = y'([itex]\frac{1}{y}[/itex]-x)
y = [itex]\frac{y'}{y}[/itex] - y'x
[itex]\frac{y'}{y}[/itex] = y+ y'x
now integrating both sides we get..
ln (y) = ∫(y + y'x dx)

But how do I integrate y+y'x?
 

Answers and Replies

  • #2
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4,806
I've come across this problem while self studying Ordinary Differential Equations and I really need help. The problem asks me to simply eliminate derivatives, I do not need to separate. The book shows the answer, but not the steps.

problem:
[itex]\frac{dy}{dx}[/itex]= [itex]\frac{y^2}{1-xy}[/itex]

answer:
ln(y)=xy

So far, I'm here...

[itex]\frac{dy}{dx}[/itex]= [itex]\frac{y^2}{1-xy}[/itex]
y' = [itex]\frac{y^2}{1-xy}[/itex]
y' = [itex]\frac{y}{\frac{1}{y}-x}[/itex]
y = y'([itex]\frac{1}{y}[/itex]-x)
y = [itex]\frac{y'}{y}[/itex] - y'x
[itex]\frac{y'}{y}[/itex] = y+ y'x
now integrating both sides we get..
ln (y) = ∫(y + y'x dx)

But how do I integrate y+y'x?
CAn't you use the product rule x'y + y'x
 
  • #3
109
2
We want to integrate.
xy must be the solution to ∫(y'x+y dx) because the derivative of xy = y'x+y(1). I just need to know how to integrate this -> ∫(y'x+y dx)
 
  • #4
798
34
We want to integrate.
xy must be the solution to ∫(y'x+y dx) because the derivative of xy = y'x+y(1). I just need to know how to integrate this -> ∫(y'x+y dx)
∫(y'x+y dx) is a non-sens.
∫(y'x+y) dx is correct.
(y'x+y) = (xy)'
∫(y'x+y) dx = ∫(yx)' dx = ? (obvious)
 
  • #5
109
2
∫(y'x+y dx) is a non-sens.

∫(y'x+y) dx is correct.

(y'x+y) = (xy)'

∫(y'x+y) dx = ∫(yx)' dx = ? (obvious)

(y'x+y) = (xy)'
Can you please explain
Thank you!
 
  • #6
11,334
4,806
(y'x+y) = (xy)'
Can you please explain
Thank you!
given two function f(x) and g(x) then the derivative of the product f*g is:

(fg)' = f'g + fg' i.e. the product rule so you look at your integral notice that it fits the right hand side and rewrite it to be

the indefinite integral: integral( (f'g+fg')dx) = integral( (fg)'dx) = fg
 
Last edited:
  • #7
109
2
Is this how you came to this conclusion?

(y'x+y)
= xy'(1+y)
= x([itex]\frac{d(1+y)}{dx}[/itex])
= x(0 + y')
= x(y')
= xy'
= (xy)'
so, ∫(xy)' dx = xy
Therefore,
∫[itex]\frac{y'}{y}[/itex] dx = ∫(xy)' dx [itex]\rightarrow[/itex] ln(y) + C = xy
Correct?
 
Last edited:
  • #8
11,334
4,806
Is this how you came to this conclusion?

(y'x+y)
= xy'(1+y)
= x([itex]\frac{d(1+y)}{dx}[/itex])
= x(0 + y')
= x(y')
= xy'
= (xy)'
so, ∫(xy)' dx = xy
Therefore,
∫[itex]\frac{y'}{y}[/itex] dx = ∫(xy)' dx [itex]\rightarrow[/itex] ln(y) + C = xy
Correct?
I think you're over complicating this. Have you learned the product rule when you took calculus 1? Just by inspection y + y'x you can see that there are two functions x and y with x' =1 and y' = y'

Hence integrating it just means you replace the y+y'x term with (xy)' then you can conclude that the integrated answer is simply y'x via the fundamental theorem of calculus right?

So is your problem going from integral(xy)' to get the ln(y) ?
 
  • #9
109
2
I think you're over complicating this. Have you learned the product rule when you took calculus 1? Just by inspection y + y'x you can see that there are two functions x and y with x' =1 and y' = y'

Hence integrating it just means you replace the y+y'x term with (xy)' then you can conclude that the integrated answer is simply y'x via the fundamental theorem of calculus right?

So is your problem going from integral(xy)' to get the ln(y) ?
How so? By using the product rule of differentation, we would be solving the equation backwards. I did not ask for the answer, I asked how to integrate y+y'x. It is clear that the derivative of xy is indeed (dy/dx)x+ y(1), but not every problem involving y in the integral will be as obvious as this problem. From the product rule of diff, sure y'(xy) = y'x+y, but I wanted to know how to integrate y'x+y, which would in fact be the way shown above, not over complicating it but simply using as a learning experience. Thanks for the help :-)
 
  • #10
109
2
Equation made exact.

[itex]{\frac{dy}{dx} = \frac{y^2}{1-xy}}[/itex]
[itex] (y^2)dx+(xy-1)dy= 0 [/itex]
[itex] \frac{∂M}{∂y} = 2y [/itex]
[itex] \frac{∂N}{∂x} = y [/itex]
[itex] 2y ≠ y[/itex]
[itex]h(y) = \frac{∂N/∂x-∂M/∂y}{M} = \frac{-1}{y}[/itex]
[itex]\mu(y) = e^∫h(y) dy[/itex]
[itex]\mu(y) = 1/y [/itex]
[itex] \frac{1}{y}[(y^2)dx+(xy-1)= 0] [/itex]
[itex] (y)dx+(x-\frac{1}{y})dy= 0 [/itex]
[itex] \frac{∂M}{∂y}= 1 = \frac{∂N}{∂x} [/itex]
[itex]\frac{∂f}{∂x} = y[/itex]
[itex] f(x,y) = xy + \phi(y) [/itex]
[itex] \frac{∂f}{∂y} = x + \phi'(y) [/itex]
[itex]x- \frac{1}{y} = x + \phi'(y) [/itex]
[itex] \phi'(y) = -1/y [/itex]
[itex] \phi(y) = -ln(y) [/itex]
[itex] f(x,y) = xy - ln y + d [/itex]
[itex] xy - ln y + d = c [/itex]
[itex] xy - ln y = \tilde{c} [/itex]
[itex] xy - ln y [/itex] [itex]\equiv 0[/itex]
[itex] ln y = xy [/itex]
 
Last edited:
  • #11
vanhees71
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How about substituting [itex]z=x y[/itex]? Then you get
[tex]z'=y+x y'.[/tex]
Then
[tex]y'=\frac{z'-y}{x}=\frac{x z'-z}{x^2} \stackrel{!}{=} \frac{y^2}{1-xy}=\frac{z^2}{x^2} \frac{1}{1-z}.[/tex]
After some algebra you find
[tex]\frac{1-z}{z}z'=\frac{1}{x},[/tex]
which is solvable by separation of variables:
[tex]\int \mathrm{d} z \frac{1-z}{z}=\int \mathrm{d} x \frac{1}{x}[/tex]
or
[tex]\ln |z|-z=\ln|x|+C \; \Rightarrow\; \ln \left |\frac{z}{x} \right| = z+C \; \Rightarrow \; \ln |y| = x y +C,[/tex]
which shows that your (implicit) solution is indeed solving the equation with the special integration constant [itex]C=0[/itex] and restricted to [itex]y>0[/itex].
 
  • #12
109
2
How about substituting [itex]z=x y[/itex]? Then you get
[tex]z'=y+x y'.[/tex]
Then
[tex]y'=\frac{z'-y}{x}=\frac{x z'-z}{x^2} \stackrel{!}{=} \frac{y^2}{1-xy}=\frac{z^2}{x^2} \frac{1}{1-z}.[/tex]
After some algebra you find
[tex]\frac{1-z}{z}z'=\frac{1}{x},[/tex]
which is solvable by separation of variables:
[tex]\int \mathrm{d} z \frac{1-z}{z}=\int \mathrm{d} x \frac{1}{x}[/tex]
or
[tex]\ln |z|-z=\ln|x|+C \; \Rightarrow\; \ln \left |\frac{z}{x} \right| = z+C \; \Rightarrow \; \ln |y| = x y +C,[/tex]
which shows that your (implicit) solution is indeed solving the equation with the special integration constant [itex]C=0[/itex] and restricted to [itex]y>0[/itex].
Is this how you would go about the alegebra for the z-sub?
[tex]y' = \frac{y^2}{1-xy} [/tex]
[tex]z = xy \Rightarrow y' = \frac{z'-y}{x} [/tex]
[tex]y'=\frac{xz'-z}{x^2} [/tex]
[tex] \frac{y^2}{1-xy}=\frac{xz'-z}{x^2}[/tex]
[tex] \frac{z^2}{x^2-zx^2}=\frac{xz'-z}{x^2}[/tex]
[tex]\frac{z^2}{1-z}= xz' - z[/tex]
[tex] \frac{z^2}{1-z} + z =\frac{xz'}{1}[/tex]
[tex] \frac{1-z}{z}=\frac{1}{xz'}[/tex]
[tex] \frac{1-z}{z} * z' =\frac{1}{x}[/tex]
*Also, must an equation be homogeneous in order for z-substitution to work? Kinda confused because I don't see any homogeneity properties and this one's clearly solvable by making this substitution.
 

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