First order DE, non homogeneous, integrating factor method

**arbolis** · February 24th 2010, 09:20 AM

I must solve $y'+t^2y=1$ . I don't think it's separable, due to the 1 on the right side of the equation.
So I used the IF method, with no success.
My attempt:
The IF is $e^{\frac{t^3}{3}}$ . Multiplying the equation by it and integrating with respect to t, I reach $e^{\frac{t^3}{3}}y=\int e^{\frac{t^3}{3}} dt$ .
I didn't know how to solve this integral and tried wolfram alpha, but the result involves the Gamma function and is pretty hard to reach I believe.
So what do I do?
Can I just say that $y=\frac{\int e^{\frac{t^3}{3}} dt}{e^{\frac{t^3}{3}}}$ ? It's not really under a nice form though. What do you think?

**HallsofIvy** · February 24th 2010, 11:48 AM

Originally Posted by arbolis

I must solve $y'+t^2y=1$ . I don't think it's separable, due to the 1 on the right side of the equation.
So I used the IF method, with no success.
My attempt:
The IF is $e^{\frac{t^3}{3}}$ . Multiplying the equation by it and integrating with respect to t, I reach $e^{\frac{t^3}{3}}y=\int e^{\frac{t^3}{3}} dt$ .
I didn't know how to solve this integral and tried wolfram alpha, but the result involves the Gamma function and is pretty hard to reach I believe.
So what do I do?
Can I just say that $y=\frac{\int e^{\frac{t^3}{3}} dt}{e^{\frac{t^3}{3}}}$ ? It's not really under a nice form though. What do you think?

That looks perfectly good to me. Although it would be slightly better not to use the same variable both inside and outside the integral. Preferable is
$y(t)= e^{t^3/3}\left(\int_0^t e^{x^3/3} dx+ C\right)$

**arbolis** · February 24th 2010, 06:05 PM

Originally Posted by HallsofIvy

That looks perfectly good to me.

Good to know!

Although it would be slightly better not to use the same variable both inside and outside the integral.

I'd love to know why.

Preferable is
$y(t)= e^{t^3/3}\left(\int_0^t e^{x^3/3} dx+ C\right)$

I think you meant $y(t)= e^{-t^3/3}\left(\int_0^t e^{x^3/3} dx+ C\right)$ .
Thanks for the help!

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