For each of the exercises given below, verify that the given func…

Question

For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

$\text{y}=\text{e}^{\text{x}}(\text{a}\cos\text{x}+\text{b}\sin\text{x})$

$\frac{\text{d}^2\text{y}}{\text{dx}^2}-2\frac{\text{dy}}{\text{dx}}+2\text{y}=0$

✓

Answer

The given differential equation is
$\text{y}=\text{e}^{\text{x}}(\text{a}\cos\text{x}+\text{b}\sin\text{x})$
$\Rightarrow\ \text{e}^{-\text{x}}\text{y}=\text{a}\cos\text{x}+\text{b}\sin\text{x}\ \ ...(1)$
Differentiating (1) twice w.r.t. .x. we get
$\text{e}^{-\text{x}}\frac{\text{dy}}{\text{dx}}+\text{y e}^{-\text{x}}(-1)=-\text{a}\sin\text{x}+\text{b}\cos\text{x}$
$\text{and}\ \ \text{e}^{-\text{x}}\ \frac{\text{d}^2\text{y}}{\text{dx}^2}+\frac{\text{dy}}{\text{dx}}\text{e}^{-\text{x}}(-1)-\Big\{-\text{y e}^{-\text{x}}+\text{e}^{-\text{x}}\ \frac{\text{dy}}{\text{dx}}\Big\}$ $=-\text{a}\cos\text{x}-\text{b}\sin\text{x}$
$ \text{or}\ \ \text{e}^{-\text{x}}\ \frac{\text{d}^2\text{y}}{\text{dx}^2}-2\text{e}^{-\text{x}}\ \frac{\text{dy}}{\text{dx}}+\text{y e}^{-\text{x}}=-\text{y x}^{-\text{x}}\ \ [\because \text{of}\ (1)]$
$ \text{or}\ \ \text{e}^{-\text{x}}\ \frac{\text{d}^2\text{y}}{\text{dx}^2}-2\text{e}^{-\text{x}}\ \frac{\text{dy}}{\text{dx}}+2\text{y e}^{-\text{x}}=0$ $\ \text{or}\ \text{e}^{-\text{x}}\Big\{\frac{\text{d}^2\text{y}}{\text{dx}^2}-2\ \frac{\text{dy}}{\text{dx}}+2\text{y}\Big\}=0$
$\ \text{or}\ \ \frac{\text{d}^2\text{y}}{\text{dx}^2}-2\ \frac{\text{dy}}{\text{dx}}+2\text{y}=0$
Hence the result.