# Thread: Differential equation question with limits

1. ## Differential equation question with limits

Sorry about the influx of my threads; panicking slightly about my distinct lack of ability when it comes to differential equations. I have the equation...

$\displaystyle y''+y'-2y=0$ with $\displaystyle y\rightarrow 0$ as $\displaystyle x\rightarrow +\infty$ and $\displaystyle y(0)=2$

I got the auxilliary equation $\displaystyle \lambda^2+\lambda -2=0$ and factorised to get $\displaystyle \lambda_1=-2 ; \lambda_2=1$, so the general solution is...

$\displaystyle y=Ae^{-2x}+Be^{x}$

But I don't know how to find A and B with those conditions I get $\displaystyle A+B=2$ but that's about as far as I got. Any ideas?

2. Originally Posted by chella182
Sorry about the influx of my threads; panicking slightly about my distinct lack of ability when it comes to differential equations. I have the equation...

$\displaystyle y''+y'-2y=0$ with $\displaystyle y\rightarrow 0$ as $\displaystyle x\rightarrow +\infty$ and $\displaystyle y(0)=2$

I got the auxilliary equation $\displaystyle \lambda^2+\lambda -2=0$ and factorised to get $\displaystyle \lambda_1=-2 ; \lambda_2=1$, so the general solution is...

$\displaystyle y=Ae^{-2x}+Be^{x}$

But I don't know how to find A and B with those conditions I get $\displaystyle A+B=2$ but that's about as far as I got. Any ideas?
check out what happens if $\displaystyle B = 0$

3. What happened to lengthy explanations on here? If B was zero then A would be 2...

4. does $\displaystyle y = 2e^{-2x}$ work ?

did you even check it out?

5. No I didn't... I'm not getting why I would've though

6. $\displaystyle y = Ae^{-2x} + Be^x$

$\displaystyle \lim_{x \to \infty} Ae^{-2x} + Be^x = 0$

what is the only value of the constant $\displaystyle B$ that will make the limit = 0 ?

7. 0... but like... urgh. Forget it. I just wanted someone to walk me through the answer step by step.

Unless you're telling me A=2 and B=0...