Unable to solve the following questionI am trying to solve the following question but I can not figure out a method to solve it. The question is:x+(y2)−(xy)dydx=y,y(12)=1

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pobijedi6wro · Accepted Answer

Step 1For x&amp;gt;0,x+y(x)2−xy(x)y′(x)=y(x) is equivalent tox+x2[(y(x)x)2−y(x)xy′(x)]=y(x)=x+x[(y(x)x)2−y(x)xy′(x)],which is equivalent toy(x)x=1+[(y(x)x)2−y(x)xy′(x)].Let z(x)=yxx, such that z(12)=2, andy′(x)=xz′(x)+z(x)z(x)=1+z(x)2−xz(x)z′(x)−z(x)2=1+−xz(x)z′(x).Step 2Since x&amp;gt;0, we have that z(x)z′(x)&amp;lt;0, but since−xz(x)z′(x)≥0, we have that 1+−xz(x)z′(x)≥1, so z(x)≥1. Therefore, it is the case that z′(x)&amp;lt;0. With this in mind, the differential equation implies[z(x)−1]2=−xz(x)z′(x),which is equivalent to−1x=z(x)[z(x)−1]2z′(x)=(1z(x)−1+1[z(x)−1]2)z′(x)=[ln⁡(z−1)−1z−1]′(x).Step 3Given the condition that x&amp;gt;0 and z(x)≥1, we have that ln⁡[z(x)−1]−1z(x)−1=−ln⁡(x)+C,which is equivalent to[z(x)−1]e−1z(x)−1=eCx.Step 4Since z(12)=2, we have that 2eC=e−1, so eC=12e. So we have that 12ex=[z(x)−1]e−1z(x)−1,which is equivalent to 1z(x)−1e1z(x)−1=2exgiven the restrictions. Here, we can use the Lambert W function, obtaining us that1z(x)−1=W(2ex),which is equivalent toz(x)=1+1W(2ex),which is equivalent to y(x)=x[1+1W(2ex)].

ruseducatives1t03 · Accepted Answer

Step 1After squaring:y2+x2−2xy=y2−(xy)dydxx−2y=−ydydx(x−2y)dx+ydy=0Step 2Let y=λx since every term is of linear degree. Then dy=xdλ+λdx.(1−2λ)dx+λ(xdλ+λdx)=(λ−1)2dx+λxdλ=0dxx=(−λ+1−1)dλ(λ−1)2ln⁡{x}=−ln⁡{|λ−1|}+1λ−1+c1x=c2e1λ−1λ−1=c2xexy−xy−xCancelling x:y−x=c2xexy−x

Unable to solve the following question I am trying

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