Question #169468

Z=a(x+log y)-x^2/2-bx

1
Expert's answer
2021-03-08T18:28:07-0500

z(x,y)=a(x+logy)x22bxz=ax+alogyx22bxz(x,y)=a(x+\log y)-\frac{x^2}{2}-bx\\ z= ax +a\log y-\frac{x^2}{2}-bx

for z(x,y):


dz=zxdx+zydydz = \dfrac{\partial z}{\partial x}dx+\dfrac{\partial z}{\partial y}dy\\

then;


zx=ax36b  ;zy=ay\dfrac{\partial z}{\partial x} = a - \frac{x^3}{6}-b\; \text{;} \qquad \dfrac{\partial z}{\partial y} = \frac{a}{y}

Let:


p=zxand q=zyp = \dfrac{\partial z}{\partial x} \qquad \text{and } \qquad q=\dfrac{\partial z}{\partial y}

then


q=ay    a=qy(i)p=ax36b(ii)q = \frac{a}{y} \implies a =qy \quad \cdots (i)\\ p = a -\frac{x^3}{6}-b \qquad \cdots(ii)

put (i) in (ii)


p=qyx36b6p=6qyx36b6p6qy+x3+6b=06(pqy)+x3+6b=06(pqy+b)+x3=0p = qy -\frac{x^3}{6}-b\\ 6p = 6qy - x^3 - 6b\\ 6p-6qy+x^3+6b=0\\ 6(p-qy)+x^3+6b=0\\ 6(p-qy+b)+x^3=0\\

The required PDE is therefore:


6(pqy+b)+x3=06(p-qy+b)+x^3=0\\


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