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I am currently solving a differential equation but I am having a little trouble figuring out my integrative factor.

I have

$$\exp\bigg({∫\frac{3}{t}dt}\bigg)$$

so to integrate I made it $\exp(\ln(t^3))$ what is the final solution? $t^3$?

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Here are the steps $$ \exp\left(\int\frac{3}{t}dt\right) = \exp\left(3\int\frac{1}{t}dt\right)$$ $$ = \exp\left(3\ln t\right) = \exp\left(\ln t^3\right) = t^3 $$ Since this is just an integrating factor, we do not need the constant.

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The answer to your confusion is that $\exp(x)=e^x$ and $\ln(x)=\log_e(x)$ are inverse functions.

This means that if you do one, and then the other, the action of the first is undone by the second. For example, $\ln e^2=2$ and $\exp(\ln(e^2))=e^2$.

You have to be careful though... you could put a negative number into $\exp(x)$ but not into $\ln x$...

Other examples of inverse functions include

  • $f(x)=2x$ and $f^{-1}(x)=\frac{x}{2}$

  • $g(x)=x^2$ and $g^{-1}(x)=\sqrt{x}$ (careful!)

  • $h(x)=x+42$ and $h^{-1}(x)=x-42$.

In your case there is an assumption that $t>0$ and when you take

$$t^3\longrightarrow \ln(t^3),$$

the action of $\exp(x)$ is to undo this and bring you back to $t^3$:

$$t^3\overset{\ln x}{\longrightarrow} \ln (t^3)\overset{\exp(x)}{\longrightarrow} t^3.$$

We have

$$\exp(\ln x)=x\,,\qquad\text{for }x>0$$ and $$\ln(\exp x)=x,$$ for any real $x$.

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