# solving a first order differential equation using separation of variables

• Aug 30th 2009, 04:22 PM
derlin
solving a first order differential equation using separation of variables
this is a rather elementary problem for diff. eq, and i solved it correctly using separation of variables, but there is a reason that the answer y=4/(x^2) is only valid for x<0.

dy/dx=y^(3/2)
dy/(y^3/2) = dx
integrate both sides to get....
-2*y^(-1/2)= x + C
-2=(x+C)*y^(1/2)
y^(1/2) = -2/(x+C)
y = 4/[(x+C)^2]

so...for some reason, when i solved for y explicitly in terms of x, I shold have been able to predict that the solution y = 4/x^2 is only valid for x<0

but I cant figure out what step in my algebra should have led me to conclude that the solution y = 4/x^2 is only valid for x<0.

any help appreciated.
• Aug 31st 2009, 01:06 AM
BobP
If $\displaystyle y^{3/2} > 0 ,$ then $\displaystyle \frac{dy}{dx} > 0$ so the solution curve will have a positive slope.

Now look at the graph of your 'solution', $\displaystyle y = 4/x^{2}$.

It has a vertical asymptote at $\displaystyle x = 0$ and its slope is negative for all $\displaystyle x>0$.
• Aug 31st 2009, 02:21 AM
chisigma
Quote:

Originally Posted by BobP
... if $\displaystyle y^{3/2} > 0 ,$ then $\displaystyle \frac{dy}{dx} > 0$ so the solution curve will have a positive slope...

Let's write the DE as...

$\displaystyle y^{'} = \sqrt {y^{3}}$ (1)

If $\displaystyle y<0$ clearly the (1) has no solution if $\displaystyle x \in \mathbb {R}$. If $\displaystyle y>0$ the derivative is expressed as 'square root' so that it can be as well positive or negative and the choice depends from the 'iniatial condition' $\displaystyle y(x_{0}) = y_{0}$ with $\displaystyle y_{0} \ge 0$. The function $\displaystyle y=\frac{4}{x^{2}}$ is then solution of (1) $\displaystyle \forall {x \ne 0}$. The (1) has another interesting properties: the value of $\displaystyle y^{'}$ doesn't depend fron $\displaystyle x$. That means that if $\displaystyle f(x)$ is solution of (1) $\displaystyle f(x+\xi)$ with $\displaystyle \xi$ arbitrary is also solution of (1), so that for each initial condition $\displaystyle y(x_{0}) = y_{0}$ with $\displaystyle y_{0} > 0$ we have two solution of (1) of the form $\displaystyle y= \frac{4}{(x - \xi)^{2}}$, one with positive slope and the other with negative slop in $\displaystyle x=x_{0}$(Rock)...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Aug 31st 2009, 05:30 AM
BobP
I would have thought that with normal usage, the RHS of the equation would be taken to be positive. However, if that is not the case aren't we looking at two different differential equations here $\displaystyle y' = +y^{3/2}$ and $\displaystyle y' = -y^{3/2}$ ?

With regards to the solution $\displaystyle y = 4/x^{2},$ (BTW $\displaystyle y = 1/x^{2}$ is not a solution), $\displaystyle dy/dx = -8/x^{3},$ so if this is to be a solution of the first of these equations it is necessary that $\displaystyle x<0$ while if it is to be a solution of the second equation we need $\displaystyle x>0.$
• Aug 31st 2009, 06:10 AM
chisigma
Thanks to your indication i was able to correct the error... $\displaystyle \frac{4}{x^{2}}$ and not $\displaystyle \frac{1}{x^{2}}$ is solution od the DE...

$\displaystyle y^{'}= \sqrt{y^{3}}$ (1)

How silly of me! (Headbang) ... The problem is therefore that the omission of the initial conditions in the usual form $\displaystyle y(x_{0})=y_{0}$ with $\displaystyle y_{0} >0$. What I have demonstrated is that with the usual approach the (1) has in general two solution satisfiyng the condition $\displaystyle y(x_{0})=y_{0}$ with $\displaystyle y_{0} >0$. As example we take $\displaystyle y(x)=\frac{4}{x^{2}}$ which is solution of the (1) with initial conditions $\displaystyle y(2)= 1$. Unfortunatley it is easy to verifiy that $\displaystyle y(x)= \frac{4}{(x-4)^{2}}$ is also solution of (1) with the same initial condition(Thinking)...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$