1. ## Substitution D.E

Ok, so i do everything to one point and can't get the right answer.
The question is:
By using the substitution y = xu solve:
$\displaystyle \frac{dy}{dx} = \frac{x^2 + y^2}{2xy}$
$\displaystyle y^2 = x(x + k)$

Now, i differentiate y = xu:
$\displaystyle \frac{dy}{dx} = u + x\frac{dy}{dx}$
After substituting in and canceling i get:
$\displaystyle u + x\frac{dy}{dx} = \frac{1 + u^2}{2u}$
I then take u to the otherside and put it into the fraction, multiply and integrate to get:
$\displaystyle -ln(1 - u^2) = lnx + c$
Taking the exponential:
$\displaystyle \frac{1}{1 - u^2} = e^lnx + c$
$\displaystyle \frac{1}{1 - u^2} = kx$
Then i substitute in y = xu ( u = y / x) :
$\displaystyle \frac{x^2}{x^2 - y^2} = kx$
Re-arranging:
$\displaystyle x^2 = kx(x^2 - y^2)$
Making y the subject does not give the anwser.

Any help would be greatly appriciated. Thank-you

2. Hello, Ashley!

$\displaystyle \frac{dy}{dx} \:=\: \frac{x^2 + y^2}{2xy}$

The answer is: .$\displaystyle y^2 \:= \:x(x + k)$

Now, i differentiate $\displaystyle y = xu\!:\;\;\frac{dy}{dx} \:=\: u + x\frac{du}{dx}$

After substituting and canceling, i get: .$\displaystyle u + x\frac{dy}{dx} \:=\: \frac{1 + u^2}{2u}$

I then take $\displaystyle u$ to the other side and put it into the fraction:

. . $\displaystyle x\frac{du}{dx} \:=\:\frac{1+u^2}{2u} - u \:=\:\frac{1-u^2}{2u} \quad\Longrightarrow\quad \frac{2u}{1-u^2}\,du \:=\:\frac{dx}{x}$

Integrate to get: .$\displaystyle -\ln(1 - u^2) \:=\: \ln x + c$
I continued this way . . .

$\displaystyle -\ln(1-u^2) \:=\:\ln x + c \quad\Longrightarrow\quad \ln(1-u^2) \:=\:-\ln x + \ln(C) \:=\:\ln\left(\frac{C}{x}\right)$

Exponentiate: .$\displaystyle 1-u^2 \:=\:\frac{C}{x}$

Back-substitute: .$\displaystyle 1 - \left(\frac{y}{x}\right)^2 \:=\:\frac{C}{x} \quad\Longrightarrow\quad \frac{x^2-y^2}{x^2}\:=\:\frac{C}{x}$

. . . . . . $\displaystyle x^2-y^2 \;=\;Cx \quad\Longrightarrow\quad y^2 \;=\;x^2 - Cx \quad\Longrightarrow\quad y^2 \;=\;x^2 + kx$

Therefore: .$\displaystyle y^2 \;=\;x(x+k)$

3. Originally Posted by Soroban
Hello, Ashley!

I continued this way . . .

$\displaystyle -\ln(1-u^2) \:=\:\ln x + c \quad\Longrightarrow\quad \ln(1-u^2) \:=\:-\ln x + \ln(C) \:=\:\ln\left(\frac{C}{x}\right)$

Exponentiate: .$\displaystyle 1-u^2 \:=\:\frac{C}{x}$

Back-substitute: .$\displaystyle 1 - \left(\frac{y}{x}\right)^2 \:=\:\frac{C}{x} \quad\Longrightarrow\quad \frac{x^2-y^2}{x^2}\:=\:\frac{C}{x}$

. . . . . . $\displaystyle x^2-y^2 \;=\;Cx \quad\Longrightarrow\quad y^2 \;=\;x^2 - Cx \quad\Longrightarrow\quad y^2 \;=\;x^2 + kx$

Therefore: .$\displaystyle y^2 \;=\;x(x+k)$

A question though, how did you get
$\displaystyle \ln(C)$

because i don't see how 'c' is equivalent to ln c ?
Or am i missing something :S?

4. Hello, Ashley!

How did you get $\displaystyle \ln(C)$

i don't see how $\displaystyle c$ is equivalent to $\displaystyle \ln c$.
Or am i missing something?
You should be familiar with this procedure by now . . .

First of all, I used two different C's.

I had an arbitrary constant, $\displaystyle c.$

Since $\displaystyle c$ is the log of some other number, $\displaystyle C$,
. . I let: .$\displaystyle c \:=\:\ln(C)$

Now why do we need something like that?

Suppose our equation came out to: .$\displaystyle \ln y \:=\:2\ln x + c$

We can exponentiate both sides: .$\displaystyle e^{\ln y} \;=\;e^{2\ln x + c}$

. . and get: .$\displaystyle y \;=\;e^{2\ln x + c}$

Technically, we're done . . . We have the answer.
But we should not be surprised that the book has simplified the answer.

Before we exponentiate, we have:

. . $\displaystyle \ln y \;=\; \underbrace{2\ln x}_{\downarrow} + \underbrace{c}_{\downarrow}$
. . $\displaystyle \ln y= \;\underbrace{\ln x^2 + \ln C}_{\downarrow}$
. . $\displaystyle \ln y \;=\;\;\;\overbrace{\ln(Cx^2)}$

Now we can exponentiate: .$\displaystyle y \;=\;Cx^2$ . . . see?

5. Originally Posted by Soroban
Hello, Ashley!

You should be familiar with this procedure by now . . .

First of all, I used two different C's.

I had an arbitrary constant, $\displaystyle c.$

Since $\displaystyle c$ is the log of some other number, $\displaystyle C$,
. . I let: .$\displaystyle c \:=\:\ln(C)$

Now why do we need something like that?

Suppose our equation came out to: .$\displaystyle \ln y \:=\:2\ln x + c$

We can exponentiate both sides: .$\displaystyle e^{\ln y} \;=\;e^{2\ln x + c}$

. . and get: .$\displaystyle y \;=\;e^{2\ln x + c}$

Technically, we're done . . . We have the answer.
But we should not be surprised that the book has simplified the answer.

Before we exponentiate, we have:

. . $\displaystyle \ln y \;=\; \underbrace{2\ln x}_{\downarrow} + \underbrace{c}_{\downarrow}$
. . $\displaystyle \ln y= \;\underbrace{\ln x^2 + \ln C}_{\downarrow}$
. . $\displaystyle \ln y \;=\;\;\;\overbrace{\ln(Cx^2)}$

Now we can exponentiate: .$\displaystyle y \;=\;Cx^2$ . . . see?

I see . We were taught it differently and not explained to us properlyy. So thankyou very much for clearing that up, much appreciated.