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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?!

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2  
First step is wrong - should be $dx/(x+4)$ on the left. –  Gerry Myerson May 3 '12 at 22:57
1  
After you do the separation of variables correctly and integrate, one "$C$" is enough. If you really really want to (but you shouldn't), a $C$ and a $D$. –  André Nicolas May 3 '12 at 23:02
    
@GerryMyerson Thanks I had a mistake, I updated my question. please have a look –  Sean87 May 3 '12 at 23:04
2  
Take the exponential of both sides. –  André Nicolas May 3 '12 at 23:07
    
@AndréNicolas No idea how to deal with that +4 on left and +1 on right :( –  Sean87 May 3 '12 at 23:09

1 Answer 1

up vote 3 down vote accepted

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

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

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

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