# What is the correct integral of this question?

In one of my course textbook's examples, we are given this DE: $\frac{dy}{dx}=x + xy$, and also that $y(0)=1$. We are asked to find its particular solution.

So, in my textbook the solution was found by the 'method of solving a linear differential equation of the first-order'.

But I decided to solve it by the 'method of separation of variables'.

So after a couple of steps, I arrived at: $$\ln|1+y| = \frac{x^2}{2} + \ln|k|$$ $$\implies \ln\left|\frac{1+y}{k}\right| = \frac{x^2}{2}$$

Now, putting the values of $x$ & $y$ we get; $k=2$. So, $$\ln\left|\frac{1+y}{2}\right| = \frac{x^2}{2}$$ $\implies$ $y = -1$ $\pm$ $2 (e)^{\dfrac{x^2}{2}}$

But according to my textbook the solution is (only) : $y = -1 + 2(e)^\dfrac{x^2}{2}$.

So did I do something wrong (And if so, what?)

• Note that you have to not only find solutions to the differential equation, but also satisfy a given initial condition – AnotherJohnDoe Oct 12 '17 at 12:52
• Yes, don't forget that $y(0) = 1$ – John Lou Oct 12 '17 at 12:53
• @JohnLou, @AnotherJohnDoe; yes I did do that- that's how I got the value of k as 2. – Mr Reality Oct 12 '17 at 12:58

## 2 Answers

$y=-1-2e^{\frac{x^2}{2}}$ is not a solution because $y(0)=-1-2e^0=-1-2*1=-3 \neq 1$

• So, if we hadn't been given that y(0)=1 then would $y= -1 - 2e^{\frac{x^2}{2}}$ be a solution too? – Mr Reality Oct 12 '17 at 13:45
• I have some questions (you do not need to respond to each one if you don't want to) :- 1) For such questions, do we have to do what you did or/and check whether all the conditions are followed in order to get rid of the solutions which are not correct?; 2) I mean my textbook didn't mention why this solution is not correct, so did it just skip the step you''ve written ( and so is it something I'm expected to do?) or was there something in this particular solution (or the 'other' solution) because of which it was obvious? Thanks! – Mr Reality Oct 12 '17 at 13:55
• I just wanted to show, in the simplest way possible, why the solution was incorrect. That doesn't mean this is the best way to answer a question (in a test or homework assignment). The best way to answer is what @Kevin did. I can't see what your textbook says, so I cannot comment on that. – Donat Pants Oct 12 '17 at 14:03
• Then, can you answer my first comment- if it was not given that y(0)=1, then both of the obtained solutions would be correct, right? – Mr Reality Oct 12 '17 at 14:14
• Not just that, the equation $$\ln|1+y| = \frac{x^2}{2} + \ln|k| =>$$ $$y = -1 + k(e)^\dfrac{x^2}{2}$$ would be correct, you can put $$k = 2$$ or $$k = -2$$ and get your 2 solutions as private cases. – Donat Pants Oct 12 '17 at 14:16

You should have, as your general solution, $$\ln|1+y|=\frac{x^2}{2}+C\ \quad\iff\quad |1+y|=e^C e^{\frac{x^2}{2}} .$$

If $1+y>0$, you have the solution $$1+y= e^Ce^{\frac{x^2}{2}}\ \quad\iff\quad y=-1+ e^Ce^{\frac{x^2}{2}} .$$

If $1+y<0$, you have the solution $$-y-1= e^Ce^{\frac{x^2}{2}}\ \quad\iff\quad -y=1+ e^Ce^{\frac{x^2}{2}} .$$

In either case, the solution can be written as $y=-1+\widehat{C}e^{\frac{x^2}{2}}$, for some constant $\widehat{C}$ (different from the $C$ above).

Placing in the initial condition then specifies the particular solution.

• Ok, this seems amazing. So am I (..more generally, is one..) expected to solve these type of problems (e.g. finding particular solutions) like you've done? Because I see that if I had done what you did, then I wouldn't have needed to even use y(0)=1 to try to find the right solution. – Mr Reality Oct 12 '17 at 14:09
• @MrReality Yes, well there are some folks who use the absolute value as a nicety to let the reader know that the $\ln$ cannot take negative values. With the absolute value though, there will always be two cases to consider. – Kevin Oct 12 '17 at 15:01