Answer to Question #348303 in Differential Equations for Farhan

Question #348303

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

1
Expert's answer
2022-06-06T14:34:03-0400
"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"


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
APPROVED BY CLIENTS