Question:
If $y=3 e^{2 x}+2 e^{3 x}$, prove that $\frac{d^{2} y}{d x^{2}}-5 \frac{d y}{d x}+6 y=0$
Solution:
Formula: -
(i) $\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{y}_{1}$ and $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\mathrm{y}_{2}$
(ii) $\frac{d\left(e^{a x}\right)}{d x}=a e^{a x}$
(iii) $\frac{\mathrm{d}}{\mathrm{dx}} \mathrm{x}^{\mathrm{n}}=\mathrm{n} \mathrm{x}^{\mathrm{n}-1}$
Given: -
$y=3 e^{2 x}+2 e^{3 x}$
$\Rightarrow \frac{d y}{d x}=6 e^{2 x}+6 e^{3 x}$
$\Rightarrow \frac{d^{2} y}{d x^{2}}=12 e^{2 x}+18 e^{3 x}$
Hence
$\Rightarrow \frac{d^{2} y}{d x^{2}}-5 \frac{d y}{d x}+6 y=6\left(2 e^{2 x}+3 e^{3 x}\right)-30\left(e^{2 x}+e^{3 x}\right)+6\left(3 e^{2 x}+2 e^{3 x}\right)$
$=0$