Question

Given Z=f(x-2y)+g(3x+y) such that Zxx + AZxy+BZyy=0 Find A and B

Answer 1

Let's denote the expression $x-2y=\zeta (x,y)$ and $3x+y=\xi(x,y)$. Then rewrite task as

(1) $Z(x,y)=f(\zeta(x,y))+g(\xi(x,y))$

The first derivatives by the rule of differentiation of nested functions will be

(2) $Z'_x=f'_{\zeta}\cdot \zeta'_x+g'_{\xi}\cdot \xi'_x=f'_{\zeta}+3g'_{\xi}$

(3) $Z'_y=f'_{\zeta}\cdot \zeta'_y+g'_{\xi}\cdot \xi'_y=-2f'_{\zeta}+g'_{\xi}$

The second derivatives are

(4) $Z''_{xx}=f''_{\zeta\zeta}\cdot \zeta'_x+3g''_{\xi\xi}\cdot \xi'_x=f''_{\zeta\zeta}+9g''_{\xi\xi}$

(5) $Z''_{xy}=f''_{\zeta\zeta}\cdot \zeta'_y+3g''_{\xi\xi}\cdot \xi'_y=-2f''_{\zeta\zeta}+3g''_{\xi\xi}$

(6) $Z''_{yy}=-2f''_{\zeta\zeta}\cdot \zeta'_y+g''_{\xi\xi}\cdot \xi'_y=4f''_{\zeta\zeta}+g''_{\xi\xi}$

Let's substitute these expressions into the equation of the problem and give similar terms.

(7) $Z''_{xx}+AZ''_{xy}+BZ''_{yy}=0\\f''_{\zeta\zeta}+9g''_{\xi\xi}+A\cdot (-2f''_{\zeta\zeta}+3g''_{\xi\xi})+B\cdot(4f''_{\zeta\zeta}+g''_{\xi\xi})=0$

(8) $f''_{\zeta\zeta}\cdot (1-2A+4B)+g''_{\xi\xi}\cdot(9+3A+B)=0$

Equation (8) can be performed for any $f$ and $g$ only if the expressions in parentheses are both 0. We get a system of two algebraic equations with two unknowns.

(9) $\begin{cases} 1-2A+4B=0 \\ 9+3A+B=0 \end{cases}$

In standard form they are

(10) $\begin{cases} 2A-4B=1 \\ 3A+B=-9 \end{cases}$

The solution to this system is easy to find $A=-\frac{35}{14}; B=-\frac{21}{14}$ or $A=-\frac{5}{2}; B=-\frac{3}{2}$

Answer: $A=-\frac{5}{2}; B=-\frac{3}{2}$