Question
If $\text{y}=\text{ae}^{2\text{x}}+\text{be}^{-\text{x}},$ show that $\frac{\text{d}^2\text{y}}{\text{dx}^2}-\frac{\text{dy}}{\text{dx}}-2\text{y}=0$
✓
Answer
Here,
$\text{y}=\text{ae}^{2\text{x}}+\text{be}^{-\text{x}}$
Differentiating w.r.t.x, we get
$\frac{\text{dy}}{\text{dx}}=2\text{a}\text{e}^{2\text{x}}$
Differentiating w.r.t.x, we get
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=4\text{a}\text{e}^{2\text{x}}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\text{a}\text{e}^{2\text{x}}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{dy}}{\text{dx}}+2\text{y}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}-\frac{\text{dy}}{\text{dx}}-2\text{y}=0$
Hence proved.
$\text{y}=\text{ae}^{2\text{x}}+\text{be}^{-\text{x}}$
Differentiating w.r.t.x, we get
$\frac{\text{dy}}{\text{dx}}=2\text{a}\text{e}^{2\text{x}}$
Differentiating w.r.t.x, we get
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=4\text{a}\text{e}^{2\text{x}}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\text{a}\text{e}^{2\text{x}}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{dy}}{\text{dx}}+2\text{y}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}-\frac{\text{dy}}{\text{dx}}-2\text{y}=0$
Hence proved.
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