y' + y = e×
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....
------ Catalog of math sources by Professor M. Maheswaran: http://uwc.edu/depts/math/resources/catalog
Woops... there were several ways to attack this type of problem here: https://www.physicsforums.com/threads/y-y-e-x-2nd-order-nonhomogenous-di...
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 work,carry on,I have got very easy method for bernulli's equation
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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....
------
Catalog of math sources by Professor M. Maheswaran:
http://uwc.edu/depts/math/resources/catalog
Woops... there were several ways to attack this type of problem here:
https://www.physicsforums.com/threads/y-y-e-x-2nd-order-nonhomogenous-di...
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 work,carry on,I have got very easy method for bernulli's equation
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