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I do not get their explanation here at all. In class, we derived the following formula for the method of integrating factors for solving first order inhomogeneous linear ODEs of the form $$y'(x)+p(x)y(x)=q(x).$$ $$y(x)=\dfrac{1}{u(x)}\left[\int u(x)q(x)dx+C\right],$$ where the integrating factor is given as $$u(x)=e^{\int p(x)dx}.$$ Following their method, I indeed got the same integrating factor $u(t)=e^{t^2}$, but with the formula above, how then am I supposed to arrive at their conclusion? Integration by parts on the $u(t)q(t)$ part seems to get me nowhere as (without using the error function), $e^{t^2}$ cannot be differentated away or integrated. Could someone perhaps help fill in the missing lines of working or explain the intermediary steps?

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    $\begingroup$ $\int u(t) q(t) \, dt=\int e^{t^2} \, 2t^3 \, dt$. Use substitution $t^2=w$ to write this as $\int e^w \, w \, dw$. Now use integration by parts. $\endgroup$
    – Anurag A
    Nov 5, 2023 at 7:55
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    $\begingroup$ The function given by this method of integrating factor is always a solution of the differential equation, whether you have a "nice" antiderivative in terms of elementary functions or not. $\endgroup$ Nov 5, 2023 at 7:56

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$$x'+2xt=2t^3$$ Integrating factor is $e^{\int2tdt}=e^{t^2}$ Multiplying I.F. both sides,

$$xe^{t^2}=\int2t^3e^{t^2}dt$$ Let $ t^2=z$ .Then $2tdt=dz$ $\therefore xe^{t^2}=\int ze^zdz =z e^z -e^z+c$ [ $c$ is integrating constant]$=(z-1)e^z+c$ $=(t^2-1) e^{t^2}+c $

$$\therefore x =t^2-1+ce^{-t^2}$$

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The method of integrating factors applies to differential equations of the form:

$$ \frac{dx}{dt} + P(t)x = Q(t) $$

which we solve by searching for a factor $I(t)$ such that:

$$ I(t)\left( \frac{dx}{dt} + P(t)x \right) = \frac{d}{dt} \left( I(t)x \right) = I(t)Q(t) $$

Such a factor can be computed to be:

$$ I(t) = e^{\int P(t) dt} $$

The solution $x(t)$ is then:

$$ x(t) = \frac{1}{I(t)} \left( \int I(t)Q(t)dt + C \right) $$

Now, in our case we have:

$$ \frac{dx}{dt} + 2 t x = 2t^3 $$

so $P(t) = 2t$ and $Q(t) = 2t^3$. The integrating factor is then:

$$ I(t) = e^{\int P(t) dt} = e^{t^2} $$

the solution is then:

$$ x(t) = e^{-t^2} \left( \int e^{t^2} \cdot 2t^3 dt + C \right) $$ $$ = e^{-t^2} \left( \int e^{t^2} d\left( \frac{t^4}{2} \right) + C \right) $$ $$ = e^{-t^2} \left( \int \frac{1}{2} e^{t^2} d\left( t^4 \right) + C\right) $$

We take $u = t^2$ and apply integration by parts:

$$ x(t) = e^{-t^2} \left( \int e^u du + C\right) $$ $$ = e^{-t^2} \left( \int u de^u + C\right) $$ $$ = e^{-t^2} \left( ue^u - \int e^u du + C\right) = e^{-t^2} \left( ue^u - e^u + C\right) $$

And so:

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

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