Question Number 191859 by mathlove last updated on 02/May/23 | ||
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Commented by mr W last updated on 03/May/23 | ||
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$${all}\:{are}\:{of}\:{type}\:{y}'+{P}\left({x}\right){y}={Q}\left({x}\right). \\ $$ | ||
Answered by qaz last updated on 02/May/23 | ||
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$$\frac{{dy}}{{dx}}+{py}={q} \\ $$$$\Rightarrow{y}={e}^{−\int{pdx}} \left({C}+\int{qe}^{\int{pdx}} {dx}\right) \\ $$ | ||