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12 Nov 2019

Find the value for k for which the DE is an exanct equation: (kx^3 y^2- e^-x) dx +( 2x^4 y +1/y)dy = 0

A. 8 B. 4 C. 1/2 D. -2 E. none

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Casey DurganLv2

22 Apr 2019

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ISBN: 9781319050733

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Related questions

can you solve this please with steps

(a) Solve the DE y 0 = sec x tan x + 2 given that y = 4 when x = Ï 4 .

(b) Solve the DE dy dx = â2x 3 + 4 given that y = 4 when x=0

Solve following differetial equations.

(1) dy/dx = 4 sin(2x)/y, y(0)=1

(2) dy/dx = -y(1+2x2)/x, y(1)=2

(3) dy/dx = -3x+4

(4) y'+2y=ex/2

(5) y'+y/5=5-x

(6) y'-2y/x=7x, y(1)=10

(7) (y2-1)dy=(t2+1)dt, y(0)=0

(8) 2(x-1)dx=(3y2+4y+y)dt, y(-1)=0

(9) (xy-3y+x-3)dy=(xy+2y-x-2)dx, y(4)=2

(10) dy=xe2ydx, y(0)=-1

(11) dy=e3x+2ydx, y(0)=1

(12) dy=y2(1-e3x)dx, y(0)=

(13) dy/(5y-2))=x dx, y(0)=10

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