For , the separated equations are
On a region containing the prescribed points, take and . Integration gives
The two specified values, verified in the original PDF, give
Dividing these relations eliminates :
The right-hand side is the square of the positive base on the left, which differs from one. Thus , and either original condition gives . Therefore
Its partial derivatives satisfy and , verifying the equation wherever its coefficients are defined. Substitution at the two assigned points gives and exactly. The formula extends to any region where and .

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