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Question

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abb0t · Answer

I used substitution for the quadratic: $$t=r^2, t^2=r^4$$ 
then i used quadratic formula. But that's where I got the wrong answer.

anonymous · Answer

So you're trying to solve the associated homogenous equation?

$$\Large \lambda^4+2\lambda^2+1=0 $$
and maybe with an additional substitution right? 

$$u=\lambda^2 $$

abb0t · Answer

Yeah.

anonymous · Answer

So 
$$\Large u=-1=\lambda^2 $$

abb0t · Answer

I got it down to where you did, used the quadratic, but i guess you get a complex complimentary solution, but for some reason I get zero.

anonymous · Answer

Yes complex it seems to me too.
$$\Large \lambda=\pm i $$

anonymous · Answer

So as you said, complex and complementary, also double roots considering the original DE.

abb0t · Answer

How did you get -1?
I get zero.

abb0t · Answer

$$\lambda ^2+2 \lambda + 1$$ right?

anonymous · Answer

$$\Large u^2+2u+1=0 $$
Compute D
$$\Large D^2=b^2-4ac=(2)^2-4\cdot1\cdot1=0 $$
so your solution is
$$\Large x_{1,2}=\frac{-b}{2a}=\frac{-2}{2}=-1 $$

abb0t · Answer

OH WAIT Nevermind, i wrote a 4 in there for some reason!!! So I was getting -2. My bad :(

anonymous · Answer

I see, but the original problem is correct?

abb0t · Answer

Yeah, it's correct. Sorry abt that. I got it from here. Thanks :)

anonymous · Answer

no problem at all, very good then ! :-)