Given $u=f(2x-y)+g(x-2y)$, show that $$2 \frac{\mathrm{d}^2 u}{\mathrm{d}x^2} + 5 \frac{\mathrm{d}^2 u}{\mathrm{d}x\,\mathrm{d}y} + 2\frac{\mathrm{d}^2 u}{\mathrm{d}y^2} = 0.$$ I'm not even sure where to start, so any help is appreciated.
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$\begingroup$By the end of the XIXth century, notations $\frac{d\,}{dx}$, $\frac{d\,}{dy}$ for partial derivatives were replaced with $\frac{\partial\,}{\partial x}$, $\frac{\partial\,}{\partial y}$. Your textbook must be very old. In case of two variables, partial derivatives are often designated by indices as well. Thus your PDE is to be rewritten in the correct form $$ 2\frac{\partial^2 u}{\partial\, x^2}+5\frac{\partial^2 u}{\partial x\partial y}+ 2\frac{\partial^2 u}{\partial\, y^2}=0\quad\text{or}\quad 2u_{xx}+5u_{xy}+2u_{yy}=0. $$ General solution of your PDE can be written in terms of two arbitrary functions $f$ and $g$ of one variable each, i.e., $u(x,y)=f(2x-y)+g(x-2y)$. To show that such $u$ does satisfy your PDE you must not forget that $f$ and $g$ are functions of just one variable. Hence $$ \frac{\partial^2 \,}{\partial\, x^2}f(2x-y)=2^2f''(2x-y)=4f''(2x-y),\\ \frac{\partial^2 \,}{\partial x\partial y}f(2x-y)=2\cdot(-1)f''(2x-y)=-2f''(2x-y),\\ \frac{\partial^2 \,}{\partial\, y^2}f(2x-y)=(-1)^2f''(2x-y)=f''(2x-y),\\ \Rightarrow 2\frac{\partial^2 \,}{\partial\, x^2}f(2x-y)+5\frac{\partial^2 \,}{\partial x\partial y}f(2x-y)+2\frac{\partial^2 \,}{\partial\, y^2}f(2x-y)=\\ =(2\cdot 4-5\cdot 2+2)f''(2x-y)=0\cdot f''(2x-y)=0. $$ Now you see that function $f(2x-y)$ does satisfy your PDE. So does the function $g(x-2y)$. Check it youself.
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