If $y=3 e^{2 x}+2 e^{3 x},$ prove that $\frac{d^{2} y}{d x^{2}}-5 \frac{d y}{d x}+6 y=0$.

Given that $y=3 e^{2 x}+2 e^{3 x}$.
First,differentiate with respect to $x$:
$\frac{d y}{d x} = 3(2)e^{2 x} + 2(3)e^{3 x} = 6e^{2 x} + 6e^{3 x}$.
Next,differentiate again with respect to $x$:
$\frac{d^{2} y}{d x^{2}} = 6(2)e^{2 x} + 6(3)e^{3 x} = 12e^{2 x} + 18e^{3 x}$.
Now,substitute these into the expression $\frac{d^{2} y}{d x^{2}}-5 \frac{d y}{d x}+6 y$:
$= (12e^{2 x} + 18e^{3 x}) - 5(6e^{2 x} + 6e^{3 x}) + 6(3e^{2 x} + 2e^{3 x})$
$= 12e^{2 x} + 18e^{3 x} - 30e^{2 x} - 30e^{3 x} + 18e^{2 x} + 12e^{3 x}$
$= (12 - 30 + 18)e^{2 x} + (18 - 30 + 12)e^{3 x}$
$= 0e^{2 x} + 0e^{3 x} = 0$.
Thus,the identity is proven.