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I have the result of a differential equation to be:


I want this to be as simplified as possible. Can it be proceeded like:


I am not sure about the equation above, but I have the final answer given to be:


Can somone please explain how this became the final answer?!

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You may need to be careful about the step before your question! It is likely that you arrived at your expression from $\frac{dx}{x+3}=3\frac{dt}{t+2}$. Integrating gets us $\ln(|x+3|)=3\ln(|t+2|)+C$. The initial condition will enable us to see how to deal with the absolute value signs, and to evaluate $C$. – André Nicolas May 17 '12 at 13:29
up vote 1 down vote accepted

From $$\ln(x+3)=3\ln(t+2)+C$$

you cannot get $$e^{(x+3)}=e^{3(t+2)+C}$$

but instead $$e^{\ln(x+3)}=e^{3\ln(t+2)+C}$$

which can be written as $$x+3=(t+2)^3 \times e^C$$ and leads to the desired result: $e^C$ is a positive constant.

If you insisted, you could write $$e^{(x+3)}=e^{\left((t+2)^3 \times e^C\right)}$$ (note the power of $3$ rather than multiplying by $3$, and multiplying by the constant rather than adding) but it would not help.

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You can imagine the constant as $\ln C'$, so you have:

$$\ln(x+3)=3\ln (t+2) + \ln C'$$

$$\ln(x+3)=\ln (t+2)^3 +\ln C'$$

$$\ln(x+3)=\ln C'(t+2)^3$$

Then, when you have natural logarithm on both sides of your equation, it means the arguments are equal (You'd write it as exponential terms otherwise):




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@MarkBennet: I think $C'$ is okay – Gigili May 17 '12 at 14:40
OK - as you now have it – Mark Bennet May 17 '12 at 15:05
@MarkBennet: I didn't change the part you said. – Gigili May 17 '12 at 15:26
No - you did it better – Mark Bennet May 17 '12 at 16:15

Write $\ln(x+3)=3\ln(t+2)+C$ as $$ x+3 = \exp(\ln (t+2)^3 + C)=\exp(\ln(t+2)^3)\cdot \exp (C).$$ But this is the same as $$x + 3 = (t+2)^3\cdot K$$ where $K = \exp(C)$.

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Why I shouldnt write the most left side as $e^{(x+3)}$ ? – Sean87 May 17 '12 at 13:11

let our C some constant $C=ln(c)$ now we have $ln(x+3)=3*lnc((t+2))$ so we have $x+3=c*(t+2)^3$ so it means that


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That's exactly what I wrote in my answer. – Gigili May 17 '12 at 14:10
And it likewise needs $c^3$ – Mark Bennet May 17 '12 at 14:11

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