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How do I solve differential equation?

$\frac{dx}{dt}=x^2+5x$ and the begining value is given $x(0)=-3$.

$\int \frac{dx}{x^2+5x}=\int dt$

$\int \frac{dx}{x^2+5x}=x(t)=-\frac{5e^{5c+5t}}{e^{5c+5t}-1}$

I get that $C$ is $\ln\frac{-3}{2}$.

Final result with constant included is that $x(t)=-\frac{5e^{5\ln\frac{-3}{2}+5t}}{e^{5\ln\frac{-3}{2}+5t}-1}$.

Is this the same as: $x(t)=-\frac{5e^{\ln\frac{-3}{2}+5t}}{e^{\ln\frac{-3}{2}+5t}-1}$. I typed this solution on my test and it's not correct.

Can someone explain me my mistake?

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  • $\begingroup$ I have seen, but the results aren't the same and the question isn't the same. Please remove comment. @E.H.E $\endgroup$
    – Alen
    Oct 15, 2016 at 22:55
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    $\begingroup$ The question is exactly the same, as is the answer. The difference is the initial condition. $\endgroup$
    – Kaynex
    Oct 15, 2016 at 22:59

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I'm not sure what you did in your integration but that looks like the mistake to me. Here is the proper continuation:

$$\int\frac{dx}{x^2+5x}=\int dt$$ $$\frac{1}{5}\log\left(\frac{x}{x+5}\right)=t+c$$ $$\frac{x}{x+5}=c\cdot e^{5t}$$ $$x(t)=\frac{5c\cdot e^{5t}}{1-c\cdot e^{5t}}$$

Using the initial condition $x(0)=-3$,

$$-3=\frac{5c}{1-c}$$ $$c=-3/2$$

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  • $\begingroup$ So my final x(t) (using constant) will be:$x(t)=\frac{\frac{-15}{2}e^{5t}}{1+\frac{5}{2}e^(5t)}$. Right @Elliot ? $\endgroup$
    – Alen
    Oct 15, 2016 at 23:07
  • $\begingroup$ Your denominator is incorrect. Just plug in $c=-3/2$ to what I have typed above. $\endgroup$
    – Elliot
    Oct 15, 2016 at 23:10
  • $\begingroup$ :$x(t)=\frac{\frac{-15}{2}e^{5t}}{1+\frac{3}{2}e^(5t)}$ now? $\endgroup$
    – Alen
    Oct 15, 2016 at 23:11

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