Answer to Question #256341 in Differential Equations for dessy

Question #256341

x(e^y)dy+((x^2)+1dx)/y=0


1
Expert's answer
2021-10-26T02:30:14-0400

Let us solve the differential equation "xe^ydy+\\frac{(x^2+1)dx}{y}=0," which is equivalent to "ye^ydy+\\frac{(x^2+1)dx}{x}=0." It follows that "\\int ye^ydy+\\int\\frac{(x^2+1)dx}{x}=C."

Let "u=y, dv=e^ydy." Then "du=dy, v=e^y."

Consequently, we get

"ye^y-\\int e^ydy+\\int (x+\\frac{1}{x})dx=C."

We conclude that the general solution of the differential equation is of the form:

"ye^y-e^y+\\frac{x^2}{2}+\\ln|x|=C."


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