2nalfq8

2022-07-12

I solved the differential equation
$\frac{dy}{dx}=\frac{x}{{x}^{2}+1}$
to get the general solution
$y=\frac{ln|x+1|+c}{2}$
Im given the initial condition
$y{y}^{\prime }-2{e}^{x}=0,y\left(0\right)=3$
but I dont know what to do with it

zlepljalz2

Expert

$y{y}^{\prime }-2{e}^{x}=0\phantom{\rule{0ex}{0ex}}y\left(0\right)=3$
$y\frac{dy}{dx}=2{e}^{x}\phantom{\rule{0ex}{0ex}}ydy=2{e}^{x}dx$
now apply integral
$\int ydy=\int 2{e}^{x}dx\phantom{\rule{0ex}{0ex}}\frac{1}{2}y\left(x{\right)}^{2}=2{e}^{x}+c\phantom{\rule{0ex}{0ex}}$
no apply y(0) to find "c" put x=0
$\frac{1}{2}y\left(0{\right)}^{2}=2{e}^{0}+c\phantom{\rule{0ex}{0ex}}\frac{1}{2}{3}^{2}=2+c$

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