# Math Help - Zill, 6th Ed., Problem 2.4.8

1. ## Zill, 6th Ed., Problem 2.4.8

I'm studying DE's on my own to fill in gaps in my knowledge. I've come across a DE which, when I follow the procedure, results in a solution that doesn't satisfy the original DE. Here's the DE:

$\dfrac{dy}{dx}=\dfrac{x+3y}{3x+y}.$

The DE is homogeneous, and either substitution ( $x=vy$ or $y=ux$) is, by symmetry, the same process. I choose $x=vy$ with $dx=v\,dy+y\,dv.$ Then the DE becomes

$(3x+y)\,dy=(x+3y)\,dx\quad\to\quad(vy+3y)\,(v\,dy+ y\,dv)=(3vy+y)\,dy\quad\to$

$v^{2}y\,dy+3yv\,dy+vy^{2}\,dv+3y^{2}\,dv=3vy\,dy+y \,dy\quad\to\quad(vy^{2}+3y^{2})\,dv=(y-v^{2}y)\,dy$

$\displaystyle\to\quad y^{2}\,(v+3)\,dv=y(1-v^{2})\,dy\quad\to\quad \frac{v+3}{1-v^{2}}\:dv=\frac{dy}{y}\quad\to\quad \ln|y|=\int\frac{v\,dv}{1-v^{2}}+3\int\frac{dv}{1-v^{2}}.$

For the first integral, let $u=1-v^{2}$, and then $du=-2v\,dv,$ or
$-\dfrac{du}{2}=v\,dv,$ and hence we get

$\displaystyle\ln|y|=-\frac{1}{2}\int\frac{du}{u}+\frac{3}{2}\,\ln\left| \frac{v+1}{v-1}\right|+C=
-\frac{1}{2}\ln|1-v^{2}|+\frac{3}{2}\,\ln\left|\frac{v+1}{v-1}\right|+C$

$\displaystyle\to\quad \ln|y^{2}|=\ln\left|\frac{(v+1)^{3}}{(v-1)^{3}}\cdot\frac{1}{1-v^{2}}\right|+C=\ln\left|\frac{(v+1)^{3}}{(v-1)^{3}}\cdot\frac{1}{(1-v)(1+v)}\right|+C=\ln\left|\frac{(v+1)^{2}}{(v-1)^{4}}\right|+C\quad\to$

$\displaystyle y^{2}=C\,\frac{(v+1)^{2}}{(v-1)^{4}}\quad\to\quad y=C\,\frac{v+1}{(v-1)^{2}}\quad\to\quad y=C\,\frac{x/y+1}{(x/y-1)^{2}}=C\,\frac{(x+y)/y}{((x-y)/y)^{2}}=\frac{x+y}{y}\,\frac{y^{2}}{(x-y)^{2}}$

$\displaystyle=\frac{y(x+y)}{(x-y)^{2}}\quad\to\quad y(x-y)^{2}=xy+y^{2}\quad\to\quad (x-y)^{2}=x+y.$

This is our implicit solution. Differentiating with respect to $x$ gives

$\displaystyle 2(x-y)(1-y')=1+y'\quad\to\quad 2(x-y)-1=y'+2y'(x-y)\quad\to\quad 2(x-y)-1=y'(x-y+1)\quad\to$

$\displaystyle y'=\frac{2x-2y-1}{x-y+1}.$

This is not the original DE, nor do I see a method for getting the original DE. I conclude that either my solution is incorrect, or my differentiation is incorrect. Where's my mistake?

2. Originally Posted by Ackbeet
I'm studying DE's on my own to fill in gaps in my knowledge. I've come across a DE which, when I follow the procedure, results in a solution that doesn't satisfy the original DE. Here's the DE:

$\dfrac{dy}{dx}=\dfrac{x+3y}{3x+y}.$

The DE is homogeneous, and either substitution ( $x=vy$ or $y=ux$) is, by symmetry, the same process. I choose $x=vy$ with $dx=v\,dy+y\,dv.$ Then the DE becomes

$(3x+y)\,dy=(x+3y)\,dx\quad\to\quad(vy+3y)\,(v\,dy+ y\,dv)=(3vy+y)\,dy\quad\to$

$v^{2}y\,dy+3yv\,dy+vy^{2}\,dv+3y^{2}\,dv=3vy\,dy+y \,dy\quad\to\quad(vy^{2}+3y^{2})\,dv=(y-v^{2}y)\,dy$

$\displaystyle\to\quad y^{2}\,(v+3)\,dv=y(1-v^{2})\,dy\quad\to\quad \frac{v+3}{1-v^{2}}\:dv=\frac{dy}{y}\quad\to\quad \ln|y|=\int\frac{v\,dv}{1-v^{2}}+3\int\frac{dv}{1-v^{2}}.$

For the first integral, let $u=1-v^{2}$, and then $du=-2v\,dv,$ or
$-\dfrac{du}{2}=v\,dv,$ and hence we get

$\displaystyle\ln|y|=-\frac{1}{2}\int\frac{du}{u}+\frac{3}{2}\,\ln\left| \frac{v+1}{v-1}\right|+C=
-\frac{1}{2}\ln|1-v^{2}|+\frac{3}{2}\,\ln\left|\frac{v+1}{v-1}\right|+C$

$\displaystyle\to\quad \ln|y^{2}|=\ln\left|\frac{(v+1)^{3}}{(v-1)^{3}}\cdot\frac{1}{1-v^{2}}\right|+C=\ln\left|\frac{(v+1)^{3}}{(v-1)^{3}}\cdot\frac{1}{(1-v)(1+v)}\right|+C=\ln\left|\frac{(v+1)^{2}}{(v-1)^{4}}\right|+C\quad\to$

$\displaystyle y^{2}=C\,\frac{(v+1)^{2}}{(v-1)^{4}}\quad\to\quad y=C\,\frac{v+1}{(v-1)^{2}}\quad\to\quad y=C\,\frac{x/y+1}{(x/y-1)^{2}}=C\,\frac{(x+y)/y}{((x-y)/y)^{2}}=\frac{x+y}{y}\,\frac{y^{2}}{(x-y)^{2}}$

$\displaystyle=\frac{y(x+y)}{(x-y)^{2}}\quad\to\quad y(x-y)^{2}=xy+y^{2}\quad\to\quad (x-y)^{2}=x+y.$

This is our implicit solution. Differentiating with respect to $x$ gives

$\displaystyle 2(x-y)(1-y')=1+y'\quad\to\quad 2(x-y)-1=y'+2y'(x-y)\quad\to\quad 2(x-y)-1=y'(x-y+1)\quad\to$

$\displaystyle y'=\frac{2x-2y-1}{x-y+1}.$

This is not the original DE, nor do I see a method for getting the original DE. I conclude that either my solution is incorrect, or my differentiation is incorrect. Where's my mistake?

I have the 9th edition. What chapter is it from?

3. It's Problem 2.4.8, so it's Problem 8 in Chapter 2.4, on "Solutions by Substitutions". Chapter 2 is "First-Order Differential Equations".

4. Why not express the second defnite integral as $\tanh^{-1} (x)$ so it's not such a mess?

I like rewriting the equation so that the substitution makes sense to me.

$\displaystyle \frac{dy}{dx} = \frac{x+3y}{3x+y} = \frac{1+3 \frac{y}{x}}{3+\frac{y}{x}}$

let $u = \frac{y}{x}$

then $\frac{dy}{dx} = x \frac{du}{dx} + u$

so $x \frac{du}{dx} + u= \frac{1+3u}{3+u}$

$x \frac{du}{dx} = \frac{1-u^{2}}{3+u}$

$\int \frac{3+u}{1-u^{2}} \ du = \int \frac{1}{x} \ dx$

$3 \int \frac{du}{1-u^{2}} \du + \int \frac{u}{1-u^{2}} \ du = \ln|x| + C$

$3 \tanh^{-1}(u) - \frac{1}{2} \ln(1-u^{2}) = \ln |x| + C$

$3 \tanh^{-1}(\frac{x}{y}) - \frac{1}{2} \ln(1 - (\frac{x}{y})^{2}) = \ln|x| + C$

I would leave it as is.

5. The substitution method is atrocious for this DE. It is an exact DE as well.

$M_y=N_x$

6. Reply to Random Variable and dwsmith:

Thank you for your additional methods to solve the problem. As elegant as your derivations may be, I don't own them. See my signature for what I mean. I'm trying to practice this substitution method, because it's one I didn't do at school. Moreover, the question I asked is what's wrong with my derivation. There has to be a mistake in it somewhere. Where is it?

7. Originally Posted by Ackbeet
Reply to Random Variable and dwsmith:

Thank you for your additional methods to solve the problem. As elegant as your derivations may be, I don't own them. See my signature for what I mean. I'm trying to practice this substitution method, because it's one I didn't do at school. Moreover, the question I asked is what's wrong with my derivation. There has to be a mistake in it somewhere. Where is it?
I understand the concept of practicing but this problem and practice on another questions.

Mathematica's solution if you would like:

8. Lol. Well, that's perfectly fine if my study methods don't suit you. What works incredibly well for me is reading through a textbook carefully, and doing every single problem as I go along.

I do have Mathematica, and it found a solution as well (looks slightly different from yours, especially in constants, so I suspect it's a different version. I have MMA 4.0). But, again, I'm pushing myself to do this problem by hand so that I understand. Mistakes are gems of learning, and I want to milk this one for all it's worth.

Thanks for looking at it.

9. Originally Posted by Ackbeet
Lol. Well, that's perfectly fine if my study methods don't suit you. What works incredibly well for me is reading through a textbook carefully, and doing every single problem as I go along.

I do have Mathematica, and it found a solution as well (looks slightly different from yours, especially in constants, so I suspect it's a different version. I have MMA 4.0). But, again, I'm pushing myself to do this problem by hand so that I understand. Mistakes are gems of learning, and I want to milk this one for all it's worth.

Thanks for looking at it.
The solution to this problem is provided on Cramster.com section 2.5 in the 9th edition still problem 8. You have to have a membership to view it though.

10. I have an account, but not of a high enough grade to view even numbered problems.

11. Originally Posted by Ackbeet
$y=C\,\frac{x/y+1}{(x/y-1)^{2}}=C\,\frac{(x+y)/y}{((x-y)/y)^{2}}=\frac{x+y}{y}\,\frac{y^{2}}{(x-y)^{2}}$
Where did $C$ go?

12. $\dfrac{dy}{dx}=\dfrac{x+3y}{3x+y}.$

I choose $x=vy$ with $dx=v\,dy+y\,dv.$ Then the DE becomes

$(3x+y)\,dy=(x+3y)\,dx\quad\to\quad(vy+3y)\,(v\,dy+ y\,dv)=(3vy+y)\,dy\quad\to$

$v^{2}y\,dy+3yv\,dy+vy^{2}\,dv+3y^{2}\,dv=3vy\,dy+y \,dy\quad\to\quad(vy^{2}+3y^{2})\,dv=(y-v^{2}y)\,dy$

$\displaystyle\to\quad y^{2}\,(v+3)\,dv=y(1-v^{2})\,dy\quad\to\quad \frac{v+3}{1-v^{2}}\:dv=\frac{dy}{y}\quad\to\quad \ln|y|=\int\frac{v\,dv}{1-v^{2}}+3\int\frac{dv}{1-v^{2}}.$

For the first integral, let $u=1-v^{2}$, and then $du=-2v\,dv,$ or
$-\dfrac{du}{2}=v\,dv,$ and hence we get

$\displaystyle\ln|y|=-\frac{1}{2}\int\frac{du}{u}+\frac{3}{2}\,\ln\left| \frac{v+1}{v-1}\right|+C=
-\frac{1}{2}\ln|1-v^{2}|+\frac{3}{2}\,\ln\left|\frac{v+1}{v-1}\right|+C$

$\displaystyle\to\quad \ln|y^{2}|=\ln\left|\frac{(v+1)^{3}}{(v-1)^{3}}\cdot\frac{1}{1-v^{2}}\right|+C=\ln\left|\frac{(v+1)^{3}}{(v-1)^{3}}\cdot\frac{1}{(1-v)(1+v)}\right|+C=\ln\left|\frac{(v+1)^{2}}{(v-1)^{4}}\right|+C\quad\to$

$\displaystyle y^{2}=C\,\frac{(v+1)^{2}}{(v-1)^{4}}\quad\to\quad y=C\,\frac{v+1}{(v-1)^{2}}\quad\to\quad y=C\,\frac{x/y+1}{(x/y-1)^{2}}=C\,\frac{(x+y)/y}{((x-y)/y)^{2}}=C\,\frac{x+y}{y}\,\frac{y^{2}}{(x-y)^{2}}$

$\displaystyle=C\,\frac{y(x+y)}{(x-y)^{2}}\quad\to\quad y(x-y)^{2}=C(xy+y^{2})\quad\to\quad (x-y)^{2}=C(x+y).$

This is our implicit solution. Differentiating with respect to $x$ gives

$\displaystyle 2(x-y)(1-y')=C(1+y')\quad\to\quad 2(x-y)-C=Cy'+2y'(x-y)\quad\to\quad 2(x-y)-C=y'(2x-2y+C)\quad\to$

$\displaystyle y'=\frac{2x-2y-C}{2x-2y+C}.$

Note: I fixed a factor of 2 missing in there as well.

Still not the original DE. More mistakes, somewhere?

13. Two things

(1) Try differentiating

$\dfrac{ (x-y)^{2}}{x+y}=C\;\;\;(*)$

$(x-y)^{2}=C(x+y),$

this will automatically eliminate C.

14. Ok, let's go:

$(x-y)^{2}=C(x+y)$

$\dfrac{(x-y)^{2}}{x+y}=C.$ Differentiating w.r.t. $x$ yields

$\dfrac{2(x+y)(x-y)(1-y')-(x-y)^{2}(1+y')}{(x+y)^{2}}=0,$ or

$2(x^{2}-y^{2})(1-y')-(x-y)^{2}(1+y')=0\quad\to$

$2(x^{2}-y^{2})-(x-y)^{2}=2(x^{2}-y^{2})y'+(x-y)^{2}y'\quad\to$

$2x^{2}-2y^{2}-(x^{2}-2xy+y^{2})=(2x^{2}-2y^{2}+x^{2}-2xy+y^{2})y'\quad\to$

$x^{2}+2xy-3y^{2}=(3x^{2}-2xy-y^{2})y'\quad\to$

$y'=\dfrac{x^{2}+2xy-3y^{2}}{3x^{2}-2xy-y^{2}}=\dfrac{x^{2}+2xy+y^{2}-4y^{2}}{4x^{2}-(x^{2}+2xy+y^{2})}
=\dfrac{(x+y)^{2}-4y^{2}}{4x^{2}-(x+y)^{2}}
=\dfrac{((x+y)-2y)((x+y)+2y)}{(2x-(x+y))(2x+(x+y))}$

$=\dfrac{(x-y)(x+3y)}{(x-y)(3x+y)}=\dfrac{x+3y}{3x+y},$

as required.

Well, Danny, you've done it again! Thanks a million. I now know something: when you're checking the answer to a DE, it's a good idea to eliminate the constant of integration. So, take the derivative in such a way as to do that. Would that be an accurate way to state it?

15. Absolutely!