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Differential Equations

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Exact Equations

DIFFEQ-YWKCLL

Solve the initial value problem:

$$(1+x)y'+y+2x=0,\ y(0)=1$$

...for $y=y(x)$

A

$y(x)=-\cfrac{x^2}{1+x}$

B

$y(x)=-\cfrac{x^2}{1+x}$

C

$y(x)=1-x$

D

$y(x)=1+x$