How to solve DE with irreducible quadratic

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SUMMARY

The discussion focuses on solving differential equations (DE) involving irreducible quadratics, specifically the equation \(\frac{dP}{dt} = -P^{2} + 10P - 26\). The recommended approach is to rewrite the DE in the form \(\frac{1}{-P^{2}+10P-26} \frac{dP}{dt} = 1\), followed by completing the square for the quadratic expression. A trigonometric substitution is suggested as a method to solve the resulting integral.

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TOD
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I'm fine solving DE with a reducible quadratic, but what happens when the quadratic is irreducible, how would I go about solving the DE? So for example,
[tex]\frac{dP}{dt}=-P^{2}+10P-26[/tex]
Thanks in advance.
 
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Try writing your DE as

[tex] \frac{1}{-P^{2}+10P-26} \frac{dP}{dt}=1[/tex]


then complete the square for [itex]-P^{2}+10P-26[/itex], and try a trig substitution.
 

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