Question #348303

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

1
Expert's answer
2022-06-06T14:34:03-0400
yy=exy2y'-y=e^xy^2




yy21y=ex\dfrac{y'}{y^2}-\dfrac{1}{y}=e^x

Use the change of variable

z=y12=y1z=y^{1-2}=y^{-1}

z=y1z=y^{-1}

Differentiate both sides with respect to xx


z=y2yz'=-y^{-2}y'

Then


y=zy2=zz2y'=-z'y^2=-z'z^{-2}

Substitute


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

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


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

exz+exz=e2xe^xz'+e^xz=-e^{2x}

d(zex)=e2xdxd(ze^x)=-e^{2x}dx

Integrate


d(zex)=e2xdx\int d(ze^x)=-\int e^{2x}dx

zex=12e2x+12Cze^x=-\dfrac{1}{2}e^{2x}+\dfrac{1}{2}C

z=12ex+12Cexz=-\dfrac{1}{2}e^x+\dfrac{1}{2}Ce^{-x}

Then


y=112ex+12Cex,y0y=\dfrac{1}{-\dfrac{1}{2}e^x+\dfrac{1}{2}Ce^{-x}}, y\not=0

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


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