xydx+{12x2(1−yln⁡y−y)}dy=0

Name: xydx+{12x2(1−yln⁡y−y)}dy=0
Uploaded: 2022-02-23T17:22:57+00:00
Description: xydx+{12x2(1−yln⁡y−y)}dy=0

Question

Gerald Lopez · Accepted Answer

Step 1  The differential equation is xydx+{12x2(1−yln⁡y−y)}dy=0  Rewrite the equation in the form of separable equation as follows.  xydx+{12x2(1−yln⁡y−y)}dy=0  xy+{12x2(1−yln⁡y−y)}dydx=0 (Divide by dx on both sides)  12x2(1−yln⁡y−y)dydx=−xy  (1−yln⁡y−y)ydydx=−2xx2  (1−yln⁡y−y)ydydx=−2x  (1−yln⁡y−y)ydy=−2xdx  Step 2   Now, integrate both sides of the obtained  equation and find the solution of the differential equation.  ∫(1−yln⁡y−y)ydy=∫−2xdx  ∫1ydy−∫ln⁡(y)dy−∫dy=−2∫1xdx  ln⁡(y)−yln⁡(y)+y−y=−2ln⁡(x)+C  ln⁡(y)−yln⁡(y)+2ln⁡(x)=C  −yln⁡(y)+ln⁡(y)+ln⁡(x2)=C

Donald Cheek · Accepted Answer

Step 1The given differntial equation is xydx+(12x2(1−yln⁡y−y))dy=0.It can be seen that it is the first order seperable differntial equation.Now seperate the variables.(−ln⁡(y)−1+1y)dy=−2xdxStep 2Integrate both sides.∫(−ln⁡(y)−1+1y)dy=∫−2xdx∫(−ln⁡(y))dy+∫(−1+1y)dy=∫−2xdx−(yln⁡(y)−∫1dy)+∫(−1)dy+∫1ydy=−∫2xdx using by part {u=ln⁡y,v′=1}−yln⁡y+∫1ydy=−∫2xdx−yln⁡y+ln⁡y=−2ln⁡x+c1−yln⁡y+ln⁡y2ln⁡x=c1−ln⁡(y)y+ln⁡(x2y)=c1 [ln⁡x+ln⁡y=ln⁡(xy)]Answer: The solution of the diffrential equation is −ln⁡(y)y+ln⁡(x2y)=c1

user_27qwe · Accepted Answer

Given: xydx+{12x2(1−yln⁡y−y)}dy=0 The eqn can be written as xydx=−12x2(1−yln⁡y−y)dy or 2x3dx=−(1y−ln⁡y−1)dy 2x3dx=(1+ln⁡y−1y)dy Integrating both sides 2∫x3dx=∫(1+ln⁡y−1y)dy+c x42=y+∫ln⁡ydy−ln⁡y+c (1) Here ∫ln⁡y=∫1⋅ln⁡y=yln⁡y−∫1y⋅ydy =yln⁡y−y(1) x22=y+yln⁡y−y−ln⁡y+c or x22=(y−1)ln⁡y+c

xydx+\{\frac{1}{2}x^{2}(1-y \ln y-y)\} dy=0

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