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# The general solution of the differential equation$e^xdy+(ye^x+2x)dx=0\; is$

$(A)\;xe^y+x^2=\;c\qquad(B)\;xe^y+y^2=\;c$$(C)\;ye^x+x^2=\;c\qquad(D)\;ye^y+x^2=\;c$

Toolbox:
• The solution for first order linear differential equation is $y.(I.F)$=Integration of Q(I.F)+C,where I.F=$e^{\int pdx}$
Step 1: