(x+2z)p+(4zx-y)q=2x²+y
full question
Solve the PDE
(x+2z)p+(4zx-y)q=2x² +y
solution :-
dxx+2z=dy4xz−y=dz2x2+y=tdx=xt+2zt,dy=−yt+4xzt,dz=2x2t+yt2x⋅dx−dy=2x2t+4xzt+yt−4xzt=(2x2+y)t⇒dz=2x⋅dx−dy⇒z=x2−y+c;Check:(x+2z)p+(4xz−y)q=(x+2x2−2y+2c)⋅2x+(4x3−4xy+4cx−y)⋅(−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;x+2zdx=4xz−ydy=2x2+ydz=tdx=xt+2zt,dy=−yt+4xzt,dz=2x2t+yt2x⋅dx−dy=2x2t+4xzt+yt−4xzt=(2x2+y)t⇒dz=2x⋅dx−dy⇒z=x2−y+c;Check:(x+2z)p+(4xz−y)q=(x+2x2−2y+2c)⋅2x+(4x3−4xy+4cx−y)⋅(−1)==2x2+y;
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