Question #263625

dy/dx=xe^x/(e^y+x^2e^y)


1
Expert's answer
2021-12-13T03:21:31-0500

dydx=xex(ey+x2ey)dydx=xexey(1+x2)eydy=xex(1+x2)dxeydy  =   xex(1+x2)dxey=ei2Ei(xi)  +   ei2Ei(x+i)  +  CEi    is   the  exponential  function    and   i     is   the  complex  number\frac{dy}{dx}=\frac{xe^x}{\left(e^y+x^{\mathrm{2}}e^y\right)} \\ \\ \frac{dy}{dx}=\frac{xe^x}{e^y\left(\mathrm{1}+x^{\mathrm{2}}\right)} \\ \\ e^ydy=\frac{xe^x}{\left(\mathrm{1}+x^{\mathrm{2}}\right)}dx \\ \\ \int{e^ydy\ \ =\ \ \ \int{\frac{xe^x}{\left(\mathrm{1}+x^{\mathrm{2}}\right)}dx}} \\ \\ e^y=\frac{e^i}{\mathrm{2}}Ei\mathrm{(}x-i\mathrm{)}\ \ +\ \ \ \frac{e^{-i}}{\mathrm{2}}Ei\mathrm{(}x+i\mathrm{)}{}{}{}{}{}\ \ +\ \ C \\ \\ Ei\ \ \ \ is\ \ \ the\ \ \mathrm{exp}onential\ \ function\ \ \ \ and\ \ \ i\ \ \ \ \ is\ \ \ the\ \ complex\ \ number \\


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