Question

from the P.D.E. z= f(x²+y²)​

Answer 1

Step-by-step explanation:

$\large \green{ \underline{ \sf \: Solution : - }}$

$\sf \: z = f( {x}^{2} + {y}^{2} )$

$\sf \: \frac{ \partial z}{ \partial x} = f'( {x}^{2} + {y}^{2} ).(2x + 0)$

$\sf \: \frac{ \partial z}{ \partial x} = f'( {x}^{2} + {y}^{2} ).(2x) \: \: \: \: \: \: \: .....(i)$

$\sf \: \frac{ \partial z}{ \partial y} = f'( {x}^{2} + {y}^{2} ).(0 + 2y)$

$\sf \: \frac{ \partial z}{ \partial x} = f'( {x}^{2} + {y}^{2} ).(2y) \: \: \: \: \: \: \: .....(ii)$

$\sf \: Divide \: eq. \: ( i) \: and (ii)$

$\sf \: \frac{ \partial z}{ \partial x} \div \frac{ \partial z}{ \partial y} = \frac{f'( {x}^{2} + {y}^{2} ).(2x)}{f'( {x}^{2} + {y}^{2} ).(2y)}$

$\sf \: \frac{ \partial z}{ \partial x} \div \frac{ \partial z}{ \partial y} = \frac{x}{y}$

$\sf \: y\frac{ \partial z}{ \partial x} = x\frac{ \partial z}{ \partial y}$

$\sf \: Required \: \: PDE$

$\green{ \boxed{\sf \: y\frac{ \partial z}{ \partial x} = x\frac{ \partial z}{ \partial y} }}$