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Take the differential equation $\dot x=\frac{dx}{dt}=x\sqrt x$.

Then $\int \frac{1}{x^{1.5}}dx=t=-2x^{-.5}+C$

Hence $x(t)=\frac{4}{(C+t)^2}$

Where if $x_0:=x(0)$, $C = \frac{2}{\sqrt x_o}$

Final Result: $$x(t)=\frac{4}{(\frac{2}{\sqrt x_o}+t)^2}$$

However, if I set $x_0$ to a positive number, such as $10$, then $x(t)$ is a decreasing function of time. This contradicts the differential equation given, because there the derivative of $x$ w.r.t $t$ is positive if $x$ is positive.

What am I doing wrong?

Edit: I've further pinpointed down where I'm making the mistake, but I still don't quite understand why it's a mistake:

Starting from the equation: $-2x^{-.5}+C=t$, I moved $C$ to the other side, intentionally leaving out the minus sign: $-2x^{-.5}=C+t$. I have always been taught that the $C$ can be any constant, so it doesn't matter whether we add a minus sign or not (because, we could always have added the constant $\tilde C:=-C $ instead).

However, here it seems to matter, because not adding the minus sign gives $x(t)=\frac{4}{(C+t)^2}$ instead of $x(t)=\frac{4}{(C-t)^2}$.

I would have quessed that this doesn't matter, because of the reason regarding $\tilde C = -C$.

However, because of the square $^2$ in the equation $x(t)=\frac{4}{(C+t)^2}$, both $\tilde C$ and $C$ reduce to $\frac {2}{\sqrt x_0}$, which would imply that $\frac {2}{\sqrt x_0}=- \frac {2}{\sqrt x_0}$.

What's going on here? what am I doing wrong?

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    Is $\dot x$ equal to $dx/dy $ or $dy/dx $? – 2017-02-25
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    i have $$\frac{1}{\sqrt{x(t)}}+\frac{1}{2}t+C=0$$ – 2017-02-25
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    I get $x(t)=\frac4{(C-t)^2}$, which is increasing for $t\lt C$ – 2017-02-25
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    @Rohan, $\dot x = \frac{dx}{dt}$ – 2017-02-25
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    Indeed $x(t)=4(C-t)^{-2}$ with $C=2x_0^{-1/2}$, that is, for every $x_0>0$, $$x(t)=\frac{4x_0}{(2-t\sqrt{x_0})^2}$$ which is defined for $$t<\frac2{\sqrt{x_0}}$$ – 2017-02-25
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    You problem is when you say "because of the square in the equation..." Because of the square, you get two values of $C$, one positive and one negative, and you need to check which one works in the original equation. Only the negative value satisfies $-2x^{-.5} = C + t$. In a simpler setting, if $a=-2$, you can infer that $a^2=4$, and it's true that from $a^2 = 4$ you can infer $a = \pm 2$, but that doesn't mean that you can infer $a=2$ from $a=-2$. – 2017-02-25

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