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OpenStudy (idealist10):
@SithsAndGiggles
OpenStudy (idealist10):
If I use Bernoulli's equation in this case, what's v?
OpenStudy (anonymous):
Another Bernoulii sub I believe. \(v=y^2\), so \(v'=2yy'\).
OpenStudy (anonymous):
You want to choose \(v\) so that it's derivative contains at least the same power of \(y\) as the term with \(y'\) in the initial equation. In this case, the \(y'\) term is \(\cdots yy'\), so you want something like \(v=\cdots y^2\). You can always manipulate the constants.
OpenStudy (idealist10):
So after I sub, I get xv'/2=3x^6+6v, right?
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OpenStudy (anonymous):
Right. The equation is linear now.
OpenStudy (anonymous):
*almost linear. You just need to do the proper rearrangement.
OpenStudy (idealist10):
xv'/2-6v=3x^6?
OpenStudy (anonymous):
Yes, and divide everything by \(\dfrac{1}{2}x\) too.
OpenStudy (idealist10):
And the integrating factor is e^-6x?
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OpenStudy (anonymous):
No the IF is something else.
\[\frac{1}{2}xv'-6v=3x^6~~\iff~~v'-\frac{12}{x}v=6x^5\]
with
\[IF=\exp\left(-\int\frac{12}{x}~dx\right)=\cdots\]