Answer to Question #348303 in Differential Equations for Farhan

Question

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

Answer 1

$$y'-y=e^xy^2$$

$$\dfrac{y'}{y^2}-\dfrac{1}{y}=e^x$$

Use the change of variable

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

$$z=y^{-1}$$

Differentiate both sides with respect to $x$

$$z'=-y^{-2}y'$$

Then

$$y'=-z'y^2=-z'z^{-2}$$

Substitute

$$z'+z=-e^x$$

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

$$u(x)=e^{\int(1)dx}=e^x$$

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

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

Integrate

$$\int d(ze^x)=-\int e^{2x}dx$$

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

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

Then

$$y=\dfrac{1}{-\dfrac{1}{2}e^x+\dfrac{1}{2}Ce^{-x}}, y\not=0$$

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