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I understand how to solve 2 element transient but am having some problems with 3+ element transients.

Specifically I'm trying to solve this equation:

$$737280 e^{-2400t}-576000 e^{-1500t} + 46080 e^{-600t}=0$$

Whenever I use WolframAlpha it always gives me a complex answer even though I know I real answer exists.

I'm also curious if a general solution to this form exists. E.g.

$$Ae^{at}+Be^{bt}+Ce^{ct}=0$$

ToBeen
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1 Answers1

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If $e^{300t}=x$, then your equation can be written

$$Ax^8+Bc^5+Cx^2=x^2(Ax^6+Bc^3+C)=0$$

Since $x>0$ for all $t$, you have just to solve

$$Ax^6+Bc^3+C=0$$

But with $y=x^3=e^{900t}$, it's just

$$Ay^2+By+C=0$$

Now you should be able to solve. Remember that $y$, which is an exponential, is always positive.

With your coefficients, the equation can even be simplified a bit more, since

$$737280y^2-576000y+46080=23040\cdot(32y^2-25y+2)$$

Jean-Claude Arbaut
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  • Yep, you beat me to it and have a solid answer anyway – David P Sep 19 '14 at 11:10
  • Wow, never thought of it like that. I gotta stop relying calculators. In what level of math is this taught (D.E., calc II, etc.)? – ToBeen Sep 19 '14 at 11:17
  • @Steven741 Well, you just have to know how to solve a trinomial. The remaining is only a matter of seeing the trick. I can't help you regarding math level, since I know only the french curriculum (trinomials are taught in high school, here). However, notice it won't work with all choices of $a,b,c$ in your second equation. – Jean-Claude Arbaut Sep 19 '14 at 11:22
  • Thanks. Trinomials are also taught in HS in the states. But there is a strong emphasis on having all the problems always be in standard quadratic form. I never even noticed this problem was a quadratic. – ToBeen Sep 19 '14 at 11:32
  • @Steven741 Ah, ok. I didn't see it either at first. When you see such an expression, it's often a good idea to try to get rid of the exponentials. Also, equations of the form $ax^{2n}+bx^n+c=0$ are really a classic. – Jean-Claude Arbaut Sep 19 '14 at 11:34