Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm trying to solve the differential equation $\frac{dy}{dx} = -xy$.

So far I've got:

$\frac{dy}{dx} = -xy$

$-xy\;dx = dy$

$\frac{1}{y}\;dy = -x\;dx$

$\ln(|y|) = -\frac{1}{2}x^2 + c$ (I combined both integration constants into one)

$y = e^c e^{-\frac{1}{2}x^2}$ or $y = -e^c e^{-\frac{1}{2}x^2}$

It looks correct, but I'm missing one particular case: $y = 0$.

$e^c$ is always positive, and thus is $-e^c$ always negative. Using both, I've included all solutions except $y = 0$.

For $y = 0$, the differential equation does hold though: $\frac{dy}{dx} = 0$ for any $x$, and $-xy = 0$ for any $x$ as well.

What am I missing in my computation?

share|improve this question
1  
You lost the case $y = 0$ when you divided by $y$. –  Qiaochu Yuan Sep 9 '11 at 18:30
1  
You divided by $y$, so of course that got rid of the case $y=0$, right? So in that stage, you could have written "either $y=0$ or ..." –  GEdgar Sep 9 '11 at 18:32
1  
You've got all but 0.00000000000001 of the solution. –  Michael Hardy Sep 9 '11 at 18:36
2  
BTW, using an asterisk for ordinary multiplication in $\TeX$ is like eating mashed potatoes by hand when silverware is available. You can write $a \times b$ or $a\cdot b$ or $ab$. –  Michael Hardy Sep 9 '11 at 18:37
    
@Michael Hardy: Thanks, I'm still trying to learn it. Learned quite a bit from your cleanup. –  pimvdb Sep 9 '11 at 18:39
show 1 more comment

2 Answers

up vote 4 down vote accepted

A "general" solution of a differential equation often misses some particular cases, which might be obtained as limits where a parameter goes to $+\infty$ or $-\infty$. In this case you would take $c \to -\infty$ so $e^c e^{-x^2/2} \to 0$. Or you could identify $\pm e^c$ as a new parameter $A$, giving you the solution $y = A e^{-x^2/2}$ (and of course the case $A=0$ works, even though it isn't a value of $\pm e^c$).

share|improve this answer
add comment

The equation is $y'+xy=0$. Usually I don't like to divide by $y$, instead, try and make this the derivative of some function by multiplying with an exponential. In this case

$$ e^{x^2/2}y'+xe^{x^2/2}y=0 \Leftrightarrow (e^{x^2/2}y)'=0$$

This means that there exists $c$ such that $e^{x^2/2}y=c$ and therefore $y=ce^{-x^2/2}$, and you didn't miss any of the cases.

Of course, the method doesn't work all the time, but this is a linear first order equation, and it is always solvable like this, and there even is a direct formula for solving $y'+p(x)y=q(x)$ when $p,q$ are continuous:

$$ \int q(x)dx \cdot e^{-\int p(x)} $$

share|improve this answer
    
And how do you know what to multiply by in situations like this? The trick is to take your suspected solution -- which will involve a constant of integration $c$ -- solve for $c$, then differentiate both sides of this equation. In this case you differentiate (both sides of) the equation $c=e^{x^2/2}y$, which gives $0=e^{x^2/2}\left[\frac{dy}{dx}+xy\right]$. So multiplying across by $e^{x^2/2}$ is what works in this case. In other situations the function may not be defined at all points $x$, or might involve $y$ which may make things a little trickier. –  Shane O Rourke Sep 9 '11 at 20:57
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.