Analysis: (If you score more, I will explain it to you! )

1, you are completely confused by the so-called "tree variable"!

2, in fact, I also hate the method of tree variables, because this is the teaching method of stupid pigs, and I can't completely blame you for being confused!

On the other hand, whether it is an independent variable or an intermediate variable is not that important at all! It is important to see who gets the derivative for whom, such as z=f(x, y)=F(ξ, ξ). If you get x, then x is the independent variable. If you get ξ, then ξ is the independent variable, but there is a differential relationship between x and ξ!

Below, I will calculate it for you in two ways. You can find that both of them can work out the result!

Method 1.

It is known that:

ξ=x+ay, η=x+by, so:

x=(aη-bξ)/(a-b)，(a≠b)

y=(ξ-η)/(a-b)，(a≠b)

Therefore:

x/？ ξ=-b/(a-b)

x/？ η=a/(a-b)

y/？ ξ= 1/(a-b)

y/？ η=- 1/(a-b)

According to the meaning of the question: u=f[x(ξ, η), y(ξ, η)], then:

u/？ ξ=f'x(？ x/？ ξ)+f'y(？ y/？ ξ)

u/？ ξ? η

=(? x/？ ξ) [f''xx(？ x/？ η)+f''xy(？ y/？ η)]+(? y/？ ξ) [f''yx(？ x/？ η)+f''yy(？ y/？ η)]

∵ The second-order partial derivatives are continuous and the mixed partial derivatives are equal.

u/？ ξ? η

=f''xx [(？ x/？ ξ) (? x/？ η)]+f''xy [(？ x/？ ξ) (? y/？ η)+(? y/？ ξ) (? x/？ η)]+f''yy [(？ y/？ ξ) (? y/？ η)]

=0

It's also VIII

4f''xx+ 12f''xy+5f''yy=0

Therefore:

(? x/？ ξ) (? x/？ η)=4

(? x/？ ξ) (? y/？ η)+(? y/？ ξ) (? x/？ η)= 12

(? y/？ ξ) (? y/？ η)=5

Do a little calculation

When a=b, the problem is obviously not satisfied!

Method 2

It is known that:

ξ=x+ay, η=x+by, so:

ξ/? x= 1，？ ξ/? y=a

η/? x= 1，？ η/? y=b

Original function: u=f(x, y), which can be converted into: u=u(ξ, η)=u[ξ(x, y), η(x, y)].

Therefore:

u/？ x=u'ξ(？ ξ/? x)+u'η(？ η/? x)

u/？ x？

=(? ξ/? x) [u''ξξ(？ ξ/? x)+u ' 'ηη(？ η/? x)]+(？ η/? x)[u ' 'ξξ(？ ξ/? x)+u''ηη(？ η/? x)]

=(? ξ/? x)？ u''ξξ+(？ η/? x)？ hello

u/？ x？ y

=(? ξ/? x) [u''ξξ(？ ξ/? y)+u ' 'ηη(？ η/? y)]+(？ η/? x)[u ' 'ξξ(？ ξ/? y)+u''ηη(？ η/? y)]

=[(? ξ/? x)(？ ξ/? y)] u''ξξ+[(？ η/? x)(？ η/? y)] u''η

u/？ y=u'ξ(？ ξ/? y)+u'η(？ η/? y)

u/？ y？

=(? ξ/? y) [u''ξξ(？ ξ/? y)+u ' 'ηη(？ η/? y)]+(？ η/? y)[u ' 'ξξ(？ ξ/? y)+u''ηη(？ η/? y)]

=(? ξ/? y)？ u''ξξ+(？ η/? y)？ hello

Therefore:

[4(? ξ/? x)？ +5(? ξ/? y)？ + 12(? ξ/? x)(？ ξ/? y)] u''ξξ+[4(？ η/? x)？ +5(? η/? y)？ + 12(? η/? x)(？ η/? y)]u ' 'η= 0

For the above equation to hold, it can only be:

4(? ξ/? x)？ +5(? ξ/? y)？ + 12(? ξ/? x)(？ ξ/? y)=0

4(? η/? x)？ +5(? η/? y)？ + 12(? η/? x)(？ η/? y)=0

Slightly calculate!