Partial Differential Equation about Rotation question

Consider the constant coefficient PDE: $$a_{11}u_{xx}+2a_{12}u_{xy}+a_{22}u_{yy}+b_{1}u_{x}+b_{2}u_{y}+cu=0$$ Show that the only ones that are unchanged under all axis-rotations (rotation invariant) have the form: $$a(u_{xx} + u_{yy}) + bu = 0$$where $a$ and $b$ are constants.

I just let $x'=x\cos A-y\sin A$ , and $y'=x\sin A+y\cos A$ ,

then get $u_{xx}\;,\;u_{xy}\;,\;u_{yy}$ in terms of $x'$ and $y'$.

But I still don't know how to get the required form since $u_{x'y'}$ doesn't vanish.

Hope someone gives me some hints or the solution.

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$$\frac{\partial u}{\partial x} = \frac{\partial x'}{\partial x} \frac{\partial u}{\partial x'} + \frac{\partial y'}{\partial x} \frac{\partial u}{\partial y'} = \cos{A} \, u_{x'} + \sin{A} \, u_{y'}$$

$$\frac{\partial u}{\partial y} = \frac{\partial x'}{\partial y} \frac{\partial u}{\partial x'} + \frac{\partial y'}{\partial y} \frac{\partial u}{\partial y'} = -\sin{A} \, u_{x'} + \cos{A} \, u_{y'}$$

Then differentiate again:

$$\frac{\partial^2 u}{\partial x^2} = \cos^2{A} \, u_{x'x'} + \sin^2{A} \, u_{y'y'} + 2 \sin{A} \cos{A} u_{x'y'}$$

$$\frac{\partial^2 u}{\partial y^2} = \sin^2{A} \, u_{x'x'} + \cos^2{A} \, u_{y'y'} - 2 \sin{A} \cos{A} u_{x'y'}$$

$$\frac{\partial^2 u}{\partial x \partial y} = -\cos{A} \sin{A} \, u_{x'x'} + (\cos^2{A}-\sin^2{A}) \, u_{x'y'} + \cos{A} \sin{A} \, u_{y'y'}$$

The coefficient of $u_{x'y'}$ obtained after plugging the above into the differential expression above is set to zero:

$$(a_{11} - a_{22}) \sin{2 A} + 2 a_{12} \cos{2 A} = 0$$

which reveals the rotation angle $A$. Note that the coefficients of $u_{x'x'}$ and $u_{y'y'}$ are not equal as you state above. The quadratic terms in the rotated coordinate system become

$$\left [ 1 + \frac{a_{12}}{\sqrt{a_{12}^2 + (a_{11}-a_{22})^2}} \right ] u_{x'x'} + \left [ 1 - \frac{a_{12}}{\sqrt{a_{12}^2 + (a_{11}-a_{22})^2}} \right ] u_{y'y'}$$

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