Let Q1=y(2) Q2 =y(3) where y=y(x) solves y'+2xy=2x^3 y(0)=1 dy/dx= (2x-y)/(x+4y) y(1)=1 y'+ycotx=y^3sin^3x y(pi/2)=1 (y^2-2x)dx+2xy dy=0 ...

Let Q1=y(2) Q2 =y(3) where y=y(x) solves

y'+2xy=2x^3 y(0)=1

dy/dx= (2x-y)/(x+4y) y(1)=1

y'+ycotx=y^3sin^3x y(pi/2)=1

(y^2-2x)dx+2xy dy=0 y(1)=1

2x^2y;+4xy=3sinx y(2pi)=0

y"+2y^-1 (y')^2=y'

Expert Solution

Colby Messinger answered 2 years ago

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