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I have the result of a differential equation to be:

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

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

$e^{(x+3)}=3e^{(t+2)+C}$

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

$x=-3+(t+2)^3C$

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 $C$.2012-05-17

4 Answers 4

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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

$x=c*(t+2)^3-3$

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    And it likewise needs $c^3$2012-05-17
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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):

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

Therefore:

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

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    No - you did it better2012-05-17
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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 $e^{(x+3)}$ ?2012-05-17
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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.