Question
If $\text{y}=\text{e}^{2\text{x}}(\text{ax}+\text{b}),$ show that $\text{y}_2-\text{4}\text{y}_1+4\text{y}=0$
✓
Answer
$\text{y}=\text{e}^{2\text{x}}(\text{ax}+\text{b}),$ Differentiating w.r.t.x, $\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{e}^{2\text{x}}(\text{a})+2(\text{ax}+b)(\text{e}^{2\text{x}})$ $\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{ae}^{2\text{x}}+2\text{y} $ Differentiating w.r.t.x,
$\Rightarrow\frac{\text{dy}}{\text{dx}}=2\text{ae}^{2\text{x}}+2\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\frac{\text{dy}}{\text{dx}}+2\text{ae}^{2\text{x}}+4\text{y}-4\text{y}=2\frac{\text{dy}}{\text{dx}}+2\frac{\text{dy}}{\text{dx}}-4\text{y}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}-4\frac{\text{dy}}{\text{dx}}+4\text{y}=0$
$\Rightarrow\text{y}_2-4\text{y}_1+4\text{y}=0$
Hence provedNeed a full question paper?
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