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For equation below:

$(t+1) \, dx=4(x+4) \, dt$

After separation I ended up with:

$(x+4)dx = \frac 4{t+1}dt $

Resulting in:

$\int x+4 \,dx = 4 \int \frac 1{t+1} \,dt$

So:

$\frac 12 x^2 + 4x + C = 4\ln(t+1) + C$

Now I have to express this as $x(t)$ and I have no clue how to. Also I am not sure if I did the above steps correctly. Any help will be appriciated!

UPDATE

As gerry pointed my mistake now I have:

$ \int \frac {1}{x+4}\,dx = 4\int \frac{1}{t+1}\,dt $

Then:

$ \ln(x+4) = 4 \ln(t+1) + C$

Still not able to express this as x(t)...how to?!

  • 0
    $e^{\ln(x+1)}=x+1$. And $e^{4\ln(t+4)}=(t+4)^4$. Get $x+1=e^C(t+4)^4$.2012-05-03

1 Answers 1

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Well, you've done all the calculus. The rest is just algebra. Taking the exponential of both sides, we get:

$x+4=e^{4\ln(t+1)+C}=e^Ce^{\ln(t+1)^4}=k(t+1)^4$

$x=k(t+1)^4-4$