What did I do wrong in solving this basic differential equation?

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SUMMARY

The discussion centers on solving the separable differential equation du/dt = (u^2)(sin t). The user correctly identifies the equation as separable and attempts to integrate it, leading to the expression 1/y = -cos t + c. However, the user mistakenly introduces a negative sign in the final solution y = -1/(cos t + c), which is incorrect. The correct approach involves recognizing that the integration of 1/(u^2) du yields -1/u, confirming that the final solution should not have a negative sign.

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Tzabcan
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I have this basic differential equation du/dt=(u^2)*(sin t)

This is obviously a separable diff eq.

So what I've done is:

g(t) = sin t h(u) = u^2

1/(u^2) du = sin t dt

Integrating both side...

1/y = - cos t + c

therefor y = - 1/(cos t + c)Which is wrong, there isn't supposed to be a minus sign apparently. I don't know what I've done wrong. Any suggestions? Thanks.
 
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Tzabcan said:
I have this basic differential equation du/dt=(u^2)*(sin t)

This is obviously a separable diff eq.

So what I've done is:

g(t) = sin t h(u) = u^2

1/(u^2) du = sin t dt

Integrating both side...

1/y = - cos t + c

\frac{d}{du}(u^{-1}) = - u^{-2}, so you should have <br /> -\frac1u = -\cos t + c<br />

therefor y = - 1/(cos t + c)Which is wrong, there isn't supposed to be a minus sign apparently. I don't know what I've done wrong. Any suggestions? Thanks.
 
pasmith said:
\frac{d}{du}(u^{-1}) = - u^{-2}, so you should have <br /> -\frac1u = -\cos t + c<br />

Ah! Wow I'm so dumb haha, 3 hours sleep :D lol.

Thanks a lot.
 

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