Question #348303

dy/dx-y= e^x y^2

Expert's answer

Use the change of variable

"z=y^{1-2}=y^{-1}""z=y^{-1}"

Differentiate both sides with respect to "x"

Then

Substitute

We get the linear equation for the function "z(x)." To solve it, we use the integrating factor:

"e^xz'+e^xz=-e^{2x}"

"d(ze^x)=-e^{2x}dx"

Integrate

"ze^x=-\\dfrac{1}{2}e^{2x}+\\dfrac{1}{2}C"

"z=-\\dfrac{1}{2}e^x+\\dfrac{1}{2}Ce^{-x}"

Then

"y=\\dfrac{2}{-e^x+Ce^{-x}}, or\\ y=0"

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