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I came across the following problem that says:

Consider the 2nd order O.D.E. $y''(x)+by'+cy=0,$ where $b,c$ are real constants.If $y=\exp(2x)$ is a solution,then which of the following is correct?

(A)$b^2+4c<0,$

(B)$b^2+4c\geq 0,$

(C)$b^2-4c<0,$

(D)$b^2-4c\geq 0.$

My Attempt: Since $y=\exp(2x)$ is a solution of the given O.D.E., it will satisfy the given O.D.E. and hence we get,$4+2b+c=0$ and now we compute $b^2-4c=b^2+c(2b+c)=(b+c)^2 \geq 0$ .option $(D)$ looks right choice.Can someone point me in the right direction? Thanks in advance for your time.

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No, $b^2+c(-2b-c)=(b-c)^2-2c^2$. –  Gerry Myerson Jan 22 '13 at 5:05
    
Sorry sir for the mistake in calculation. –  learner Jan 22 '13 at 5:07
    
OK, so computing $b^2+4c$ got you nowhere. Still, it was a good idea. Now try computing $b^2-4c$. –  Gerry Myerson Jan 22 '13 at 5:16
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Thanks a lot sir for the feedback.I have got it. –  learner Jan 22 '13 at 7:09
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3 Answers 3

up vote 2 down vote accepted

You can do it via another way. We kow that for any linear OE with constant coefficient, there is an auxiliary equation. This equation for the OE above is $$m^2+bm+c=0,~~~(1)$$ If $y=\exp(2x)$ be one solution so $m=2$ satisfy the equation above and so we have $$4+2b+c=0$$ which is as you got already, and that the equation (1) has solutions. These solutions may be different $m_1\neq m_2$, and may be happen twice, $m_1=m_2=2$. In any cases, D looks correct.

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Nice alternative route! +1 –  amWhy Feb 11 '13 at 0:04
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If $y=e^{2x}$ is a solution, $2$ is a root of the characteristic polynomial $r^2+br+c$. If the polynomial has real roots, its discriminant, $b^2-4c$, should be greater than or equal to $0$. This corresponds to answer D.

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You have the solution $y(x)=e^{2x}$ which means you have one of the roots $m=2$ of the auxiliary equation

$$ m^2+bm+c=0, $$

of the ode. The discriminant of the equation is

$$ b^2-4c. $$

Since you have one real solution $m=2$, then the other root is real too and $b^2-4c \geq 0. $

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