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Sagot :
Yes, y = e^xy is an implicit solution of the differential equation
(1-xy)y'=y^2
Here,
The differential equation (1-xy)y' = y^2
And, y = e^xy is an implicit solution of the differential equation
(1-xy)y'=y^2.
What is Differential equation?
A differential equation is a mathematical equation that relates some function with its derivatives.
Now,
To show y = e^xy is an implicit solution of the differential equation
(1-xy)y'=y^2, we have to find the solution of differential equation.
The differential equation is;
[tex](1-xy)y'=y^2\\\\\\\\[/tex]
[tex](1-xy) \frac{dy}{dx} = y^2[/tex]
[tex]\frac{dx}{dy} + \frac{x}{y} = \frac{1}{y^{2} }[/tex]
It is form of [tex]\frac{dx}{dy} + Px = Qy[/tex],
Where, P is the function of y and Q is the function of x.
Hence, Integrating factor = [tex]e^{\int\limits {\frac{1}{y} } \, dy} = e^{lny} = y[/tex]
The solution is;
[tex]x y = \int\limits {\frac{1}{y^{2} }y } \, dy + c[/tex]
[tex]xy = lny + c[/tex]
Take c = 0, we get;
[tex]xy = lny\\\\y = e^{xy}[/tex]
Hence, y = e^xy is an implicit solution of the differential equation
(1-xy)y'=y^2
Learn more about the differential equation visit:
https://brainly.com/question/1164377
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