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Integrating factor Q

Solve the initial-value problem

dx/dt=6x-1 x(0)=3

I have the solution to the problem but I don't understand one of the steps in the problem ;

dx/dt -6x=-1

(dx/dt - 6x ) (e^-6t) = -e^-6t - multiplied through by the integrating factor of e^-6t

This is simply
d/dt ( (e^-6t)(x))=-e^-6t

I don't get the above step , could anyone explain it . I was thinking maybe of using the substitution u=e^-6t

Thanks for any help .

Reply 1

Notnek

By the product rule,

\displaystyle \frac{d}{dt}\left(e^{-6t}x\right) = (-6e^{-6t} \times x) + (e^{-6t}\times \frac{dx}{dt}) =...

Does that makes sense? This step is the same for any integrating factor question and after doing a few, you'll notice that it's the same each time and you should get the hang of it.

(edited 13 years ago)

Reply 2

Rubgish

When using an integrating factor, the idea is to try and get it back into a product rule form. Assuming you've found the integrating factor correctly, then the LHS of your equation should always be the expanded form of the product rule. If you then put it back into the product form, it makes it much easier to integrate than it would have been otherwise.