Bernoulis
y' + y = e×
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(y'-y=0) are y=C(x)*exp(x) .
Hi,
I'll try)
(y'-y=0) are y=C(x)*exp(x) . Substitute in the equation with the right part, then we get : C'(x)*exp(x) =e^x, from c=x+a. So, y=(x+a)*e^x, where a= const....
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Woops... there were several
Woops... there were several ways to attack this type of problem here:
https://www.physicsforums.com/threads/y-y-e-x-2nd-order-nonhomogenous-d…
so now I don't know what is really right...
Maybe try some solvers:
https://www.symbolab.com/solver/solid-geometry-calculator
http://yourhomeworkhelp.org/do-my-geometry-homework/
Thank you so much for your
Thank you so much for your work,carry on,I have got very easy method for bernulli's equation