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This is a 3 part question (I seem to have found (a), but I am not confident about (b) and thus (c) as well):

(a) if y is a solution, show that v:= $yx^{-1}$ satisfies the differential equation:

$v'=a \sqrt{1+v^2}$

My solution is as follows:

$ v=\dfrac{y}{x}$

$v'=\dfrac{d}{dx}(\dfrac{y}{x})$

$ v'= \dfrac{y'}{x}-\dfrac{y}{x^2}$

Now substituting y', I have:

$v' = \dfrac{1}{x}(\dfrac{y}{x}+a\sqrt{x^2+y^2})-\dfrac{y}{x^2} $

Distributing the $\dfrac{1}{x}$, factoring out $x^2$ from the square root, and then substituting for v I get:

$v' = \dfrac{y}{x^2}+\dfrac{ax\sqrt{1+v^2})}{x}-\dfrac{y}{x^2} $

Cancelling gives:

$v'=a \sqrt{1+v^2}$

Does this sufficiently answer this question? (Note: This is a 2 point question)

(b) Use the differential equation in part (a) to get the general solution to:

$ xy' = y + ax \sqrt{x^2+y^2} $

This is a 3 point question and where I am having a little trouble, my strategy has been as follows:

Since $ v = \dfrac{y}{x}$, then:

$\int v'\,dv = a\int \sqrt{1+v^2}\,dv = \dfrac{y}{x}$

Substituting $v = sinhx$, (I am not sure if I can do this, or have used this correctly) I get:

$v= a\int coshx\, dx = asinhx+c=\dfrac{y}{x}$

Therefore, $y = axsinhx+c$

Now since $y'= \dfrac{y}{x}+a\sqrt{x^2 + y^2}$

substituting $y = axsinhx+c$, and pulling the constant out of the square root and simply adding it to the end, I have just observed and tried out that when $a=1$,

$y = (1)xsinhx+c$ seems to be a valid solution.

But I just guessed $a$, so I need a bit of guidance to give a solid answer to this.

(c) Find the particular solution of $ xy' = y + ax \sqrt{x^2+y^2} $ with $y(1)=0$

Perhaps trivially:

$0 = (1)sinh(1) + c$, gives $c=-sinh(1)$

Thus my final answer is: $y = (1)xsinhx-sinh(1)$

Any help/guidance on this would be greatly appreciated!

1 Answers 1

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After getting differential eqn $$\frac{dv}{dx}=a\sqrt{1+v^2}$$ $$\implies \int\frac{dv}{\sqrt{1+v^2}} =\int a dx$$ $$\implies \ln(v+\sqrt{1+v^2})=ax+c$$

Putting $v=\frac{y}{x}$ gives $$\ln(\frac{y+\sqrt{x^2+y^2}}{x})=ax+c$$

Since $y(1)=0\implies \ln(1)=a+c\implies c=-a$ (as $\ln(1 )=0$)

Thus, particular solution is $$ \ln(\frac{y+\sqrt{x^2+y^2}}{x})=a(x-1)$$

EDIT:

If you want $y$ explicitly, then, $$\frac{y+\sqrt{x^2+y^2}}{x}=e^{a(x-1)}\implies {y+\sqrt{x^2+y^2}}=xe^{a(x-1)}$$ $$\sqrt{x^2+y^2}=xe^{a(x-1)}-y$$

Squaring both sides,

$$x^2+y^2=x^2e^{2a(x-1)}+y^2-2xye^{a(x-1)}$$ $$\implies 2xye^{a(x-1)}=x^2(e^{2a(x-1)}-1)$$ $$\implies y=x(\frac{e^{a(x-1)}-e^{-a(x-1)}}{2})$$

But squaring usually changes the domain, so be careful.

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    Hi @Avatar - Thanks for your help. So would you say it is not possible to express the answer to part (b) in terms of y only? Also, my fluke answer seems to work, and satisfies the requires of the differential equation - is this possible? Or have I done something wrong? Or perhaps, I have given a particular value for the constant a, and shouldn't have?2012-09-25
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    $v=\frac{y}{x}$, so when you are substituting $v=\sinh x$, there is a problem , you can't use the same variable $x$ and so you can't integrate $\int \sqrt{1+v^2}dx$2012-09-25
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    I knew something wasn't right with that. Thanks for clearing that up for me. Am I right to say the solution to part (b) cannot be solved for y explicitly?2012-09-25
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    i edited my answer. check it.2012-09-25
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    In this case, since it is given that $x>0$, squaring should be fine?2012-09-25
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    right!, there is no problem in squaring here2012-09-25
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    Hi Avatar, just noticed there seems to be a missing x in your last line. There should be an x left over?2012-09-25