Question #192876

(x+2z)p+(4zx-y)q=2x²+y


1
Expert's answer
2021-05-13T18:15:39-0400

full question

Solve the PDE 

(x+2z)p+(4zx-y)q=2x² +y


solution :-

dxx+2z=dy4xzy=dz2x2+y=tdx=xt+2zt,dy=yt+4xzt,dz=2x2t+yt2xdxdy=2x2t+4xzt+yt4xzt=(2x2+y)tdz=2xdxdyz=x2y+c;Check:(x+2z)p+(4xzy)q=(x+2x22y+2c)2x+(4x34xy+4cxy)(1)==2x2+y;\frac{dx}{x+2z}=\frac{dy}{4xz-y}=\frac{dz}{2x^2+y}=t\\ dx=xt+2zt,\quad dy=-yt+4xzt,\quad dz=2x^2t+yt\\ 2x\cdot dx-dy=2x^2t+4xzt+yt-4xzt=(2x^2+y)t\Rightarrow\\ dz=2x\cdot dx-dy \Rightarrow z=x^2-y+c;\\ Check:\\ (x+2z)p+(4xz-y)q=(x+2x^2-2y+2c)\cdot 2x+(4x^3-4xy+4cx-y)\cdot(-1)=\\ =2x^2+y;


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