Question

Exercise : 1
Form the partial differential equation by eliminating the arbitrary constant from
the following equations
1.
z= ax +by+a2+ b2​

Answer 1

Answer:

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Answer 2

Basic Notations Used :-

$\boxed{ \bf \: \dfrac{\partial \:z }{\partial \:x} = p}$

$\boxed{ \bf \: \dfrac{\partial \:z }{\partial \:y} = q}$

Let's solve the problem now!!

$\rm :\longmapsto\:z = ax + by + {a}^{2} + {b}^{2} - - - (1)$

On differentiating partially w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{\partial \:z}{\partial \:x} = a$

$\rm :\implies\:\boxed{ \bf \: a = p} - - - (2)$

On differentiating equation (1) partially w. r. t. y, we get

$\rm :\longmapsto\:\dfrac{\partial \:z}{\partial \:y} = b$

$\rm :\implies\:\boxed{ \bf \: b = q} - - - (3)$

Now,

On substituting the values of a and b in equation (1), we get

$\boxed{ \rm \: z = px + qy + {p}^{2} + {q}^{2} \: is \: required \: differential \: equation} \:$

Additional Information :-

$\boxed{ \bf \: \dfrac{ {\partial}^{2}z }{ \partial \: y{\partial \: }x} = \dfrac{ {\partial}^{2}z }{ \partial \: x{\partial \: }y} = s}$

$\boxed{ \bf \: \dfrac{ {\partial \:}^{2}z }{ {\partial \:x}^{2} } = r}$

$\boxed{ \bf \: \dfrac{ {\partial \:}^{2}z }{ {\partial \:y}^{2} } = t}$

Remark :-

If the number of arbitrary constants to be eliminated from the given equation is less than or equal to the number of independent variables, then there exists partial differential equations of order one. i.e in terms of p and q only

If the number of arbitrary constants to be eliminated in the given equatoon is greater than the number of independent variables, then there exists partial differential equations of order two or of higher order, i.e in terms of p, q, r, s, t.