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from solution to mother equation
- Thread starter naviakam
- Start date
thanks
how the exponential equation is differentiated?
You should be excited to see the exponential; natural logarithms are again promising because they turn multiplication of inputs into addition of outputs, we want [MATH]k[/MATH] to be added rather than multiplied so that the differentiation wipes it out, and this natural log can cancel out the exponential quite nicely! I would start by moving [MATH]y_0[/MATH] to the left and taking the natural log of both sides.
You should be excited to see the exponential; natural logarithms are again promising because they turn multiplication of inputs into addition of outputs, we want [MATH]k[/MATH] to be added rather than multiplied so that the differentiation wipes it out, and this natural log can cancel out the exponential quite nicely! I would start by moving [MATH]y_0[/MATH] to the left and taking the natural log of both sides.
Y=P(p)Y1(x)+(1−P(p))Y2(x), where P(6mbar)=0 and P(10mbar)=1.
How to calculate Y(V) and E(V)?
Potential is V=YE/It
where t is the time, E energy, Y intensity, V potential, I current.
Last edited:
HallsofIvy
Elite Member
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- Jan 27, 2012
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That makes no sense. You have Y defined in terms of variables x and p while V is defined in terms of variable t.
That makes no sense. You have Y defined in terms of variables x and p while V is defined in terms of variable t.
Should be redefined as follows:
Y=P(p)Y1(E)+(1−P(p))Y2(E), where P(6mbar)=0 and P(10mbar)=1.
How to calculate Y(V) and E(V)?
Potential is V=YE/It
I and t constant.