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OpenStudy (lgbasallote):
this is linear right?
OpenStudy (anonymous):
for whatever reason I wanna take the derivative but that's not what I'm supposed to do . Yes linear
OpenStudy (lgbasallote):
\[\LARGE \frac{dy}{dx} - 5y = x\]
use integrating factor
\[e^{\int -5dx} = e^{-5x}\]
then do that \[y(I.F.) = \int Q(I.F.)dx\] thingy
\[\large y(e^{-5x}) = \int xe^{-5x}dx\]
where did you get confused?
OpenStudy (anonymous):
ok I have never done this before
OpenStudy (anonymous):
e?
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OpenStudy (anonymous):
ok lets see what's going on... we have \[e^{\int -5dx}\] and \[e^{ -5x}\]
OpenStudy (lgbasallote):
oh you havent? it's just \[\frac{dy}{dx} + Py = Q\] where P and Q are functions of x then you solve for integrating factor
\[e^{\int Pdx}\]
once you have solved the integrating factor, you use the equation
\[y(I.F.) = \int Q(I.F.)dx\]
where I.F. means integrating factor
OpenStudy (lgbasallote):
are you not familiar with integrating factors in linear DE?
OpenStudy (anonymous):
I'm not sure, P in this case is 5x
OpenStudy (lgbasallote):
no it's just 5
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OpenStudy (anonymous):
ok
OpenStudy (lgbasallote):
wait it's -5
OpenStudy (anonymous):
how about Q 1?
OpenStudy (lgbasallote):
Q would be x
\[\large \frac{dy}{dx} + Py = Q \implies \frac{dy}{dx} - 5y = x\]
OpenStudy (anonymous):
ok
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