Show z=xf(x+y)+g(x+y) solves r-2s+t=0 (with r=z_(xx)…

Show $z=xf(x+y)+g(x+y)$ solves $r-2s+t=0$ (with $r=z_{xx}$, $s=z_{xy}$, $t=z_{yy}$); find the solution with $z|_{x=0}=ay^2$, $z|_{y=0}=bx^2$.

Step 1 · Verify the ansatz

With $u=x+y$: $r=2f'+xf''+g''$, $s=f'+xf''+g''$, $t=xf''+g''$, so \[ r-2s+t = (2f'-2f') + (xf''-2xf''+xf'') + (g''-2g''+g'') = 0. \]

Step 2 · Apply $z|_{x=0}=ay^2$

$g(y)=ay^2$, so $g(v)=av^2$ for all $v$.

Step 3 · Apply $z|_{y=0}=bx^2$

$xf(x)+g(x)=bx^2 \Rightarrow xf(x)=(b-a)x^2 \Rightarrow f(x)=(b-a)x$, so $f(v)=(b-a)v$.

Step 4 · Assemble and simplify

\[ z = x(b-a)(x+y)+a(x+y)^2 = bx^2+(a+b)xy+ay^2. \]

Verification

$r=2b$, $s=a+b$, $t=2a$: $r-2s+t=2b-2(a+b)+2a=0$. \checkmark $z|_{x=0}=ay^2$, $z|_{y=0}=bx^2$. \checkmark