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I'm trying to find the general solution of the following first order differential equation (using the integrating factor method): $$\frac{dx}{dt} - 2tx=t^3$$

I found the integrating factor to be $\mathrm{e}^{-t^2}$ meaning I would have to integrate $$\frac{t^3}{\mathrm{e}^{-t^2}}$$ but I don't know how to integrate this. can anyone help?

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3 Answers 3

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Your integrating factor is correct, but your integral is off. Multiplying by the integrating factor yields

$$e^{-t^2}x'-2te^{-t^2}x=(e^{-t^2}x)'=t^3e^{-t^2}$$

$$e^{-t^2}x=\int t^3e^{-t^2}dt$$

$$u=t^2,du=2tdt$$

$$\int t^3e^{-t^2}dt=\frac12\int ue^{-u}du=$$

$$-\frac12ue^{-u}+\frac12\int e^{-u}du=-\frac12e^{-u}(u+1)$$

$$e^{-t^2}x=-\frac12e^{-t^2}(t^2+1)+C$$

$$x=-\frac12(t^2+1)+Ce^{t^2}$$

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$$\int \frac{t^3}{e^{-t^2}}dt = \int t^3e^{t^2}dt$$ Let $u=t^2\implies du=2tdt$

Now we have: $$\frac{1}{2}\int t^2e^udu=\frac{1}{2}\int ue^udu$$

Now integrate by parts.

If you need further help, just leave a comment.

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It should be noted that I did not check the DE's portion to see if that was done correctly. Rather, I only describe how to integrate the above fraction... –  anorton Feb 20 '13 at 16:19

The idea of the integrating factor comes from an observation that $$ t^3 = \frac{\mathrm{d} x(t)}{\mathrm{d} t} - 2 t x(t) = \mathrm{e}^{t^2} \frac{\mathrm{d}}{\mathrm{d} t} \left( \mathrm{e}^{-t^2} x(t) \right) $$ Thus the solution to the equation is $$ \mathrm{e}^{-t^2} x(t) = x_0 + \int_0^t \mathrm{e}^{-s^2} s^3 \mathrm{d} s = x_0 + \int_0^t \mathrm{e}^{-s^2} s^2 \mathrm{d} \left( \frac{s^2}{2} \right) \stackrel{u = s^2}{=} x_0 + \frac{1}{2} \int_0^{t^2} \mathrm{e}^{-u} u \mathrm{d} u $$ The latter integral can be integrated by parts, giving: $$ x(t) = \left(x_0 + \frac{1}{2} \right) \mathrm{e}^{t^2} - \frac{1+t^2}{2} $$

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