Wroian Differential Equation.3y″+(6x)y′+3exy=0 and two y1,y2 are two partial solutions of such that W(y1,y2)≠0.(where W(y1,y2)=W(x) is the Wroian of y1,y2). If it is known that W(1)=2, calculate W(10). I know that i have to find a function first but i do not know how. Also i can check if they are linearly independent but how can i calculate W(10). I can not find any solved examples.

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lilaznkid54rcz · Accepted Answer

Step 1The Wroian of y1 and y2 is defined asW(y1,y2)=det[y1y2y1′y2′]=y1y2′−y2y1′;we may easily find the derivativeW′=y1′y2′+y1y2″−y2′y1′−y2y1″=y1y2″−y2y1″;further progress is made using the given equation3y″+(6/x)y′+3exy=0,out of which the constant factor 3 may be divided, leavingy″+(2/x)y′+exy=0,which we know y1 and y2 solve; thus we have=y1(−((2/x)y2′+exy2))−y2(−((2/x)y1′+exy1))=−(2/x)y1y2′−exy1y2+(2/x)y2y1′+exy1y2=−(2/x)y1y2′+(2/x)y2y1′=−(2/x)(y1y2′−y2y1′)=−(2/x)W;once the clutter of this equation is removed we are left withStep 2W′=−(2/x)W,which is a case of Abel&#039;s identity; the unique solution to (6) taking the value W(1) at 1 isW(x)=W(1)exp⁡(−∫1x(2/s)ds);we may easily evaluate the integral:∫1x(2/s)ds=2∫1x(1/s)ds=2(ln⁡x−ln⁡1)=2ln⁡x=ln⁡x2;Therefore, W(x)=W(1)exp⁡(−ln⁡x2)=W(1)exp⁡(ln⁡x−2)=W(1)x−2.Now with W(1)=2,x=10,W(x)=2(10)−2=2(.01)=.02.

Wroian Differential Equation. 3y''+(6/x)y'+3e^xy = 0 and two y_{1}, y_{2} are two

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