# $y''(x)+by'+cy=0,$ where $b,c$ are real constants

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
Thanks a lot sir for the feedback.I have got it. – learner Jan 22 '13 at 7:09

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

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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