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I stacked about an equation $\frac{dp}{dt}=t^2p-p+t^2-1$ I know this equation will be $p=$(something about t) I got $$\int\frac{1}{p+1}dp=\int(t^2-1)dt$$

After take the integral I got

$$\ln|p+1|+c = \frac{1}{3}t^2-t+c$$

after this step I stacked. How can I simplify like p = something? Is it possible to write $e^{p+1}=\frac{1}{3}t^2-t+K$ and rewrite like $(p+1)\ln e=\ln(\frac{1}{3}t^2-t+K)$?

If you know how to solve this question could you post your idea on the wall?

Thank you!

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$\int (t^2 -1)dt = \frac{1}{3} t^3 - t$. Typo probably. – Hidde Feb 28 '12 at 6:39
up vote 3 down vote accepted

I assume that you meant to write $t^2p$ and not $t^2x$? If so, then you're right to conclude $\ln|p+1|+c_1=\frac{1}{3}t^3-t+c_2$, which you can rewrite as $\ln|p+1|=\frac{1}{3}t^3-t+c$ since $c_2-c_1=c$ is an arbitrary constant due to both $c_1$ and $c_2$ being arbitrary. Now raise each side to the power of $e$ to conclude $|p+1|=\exp(\frac{1}{3}t^3-t+c)$, so that your solutions are $p(t)=\exp(\frac{1}{3}t^3-t+c)-1$ and $p(t)=-\exp(\frac{1}{3}t^3-t+c)-1.$

If you like, you can rewrite $\exp(\frac{1}{3}t^3-t+c)$ as $e^{\frac{1}{3}t^3-t}e^c$; since $e^c$ is an arbitrary positive number, you can rewrite your general solution as $p(t)=Ce^{\frac{1}{3}t^3-t}-1$ for an arbitrary nonzero number $C.$

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Ah, $t^2p$! That makes sense! Wish I had thought of that as what Ryu really meant before I posted my own answer! – Gerry Myerson Feb 28 '12 at 6:47

You "stacked" a lot earlier than you think you did. There is no (correct) way to get from $${dp\over dt}=t^2x-p+t^2-1$$ to $$\int{1\over p+1}\,dp=\int(t^2-1)\,dt$$ To begin with, $x$ seems to have disappeared; but even if it hadn't been there in the first place, you've made a mistake in the algebra when you attempted to divide through by $p+1$.

Rather, if the $x$ isn't there, rewrite as $${dp\over dt}+p=t^2+t-1$$ then recognize this as a first-order linear differential equation: have you learned a technique for solving those?

And, if the $x$ really is there, what does it mean?

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