Question
If $\text{y}=\text{e}^{2\text{x}}(\text{ax}+\text{b}),$ show that $\text{y}_2-\text{4}\text{y}_1+4\text{y}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=2\text{ae}^{2\text{x}}+2\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\frac{\text{dy}}{\text{dx}}+2\text{ae}^{2\text{x}}+4\text{y}-4\text{y}=2\frac{\text{dy}}{\text{dx}}+2\frac{\text{dy}}{\text{dx}}-4\text{y}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}-4\frac{\text{dy}}{\text{dx}}+4\text{y}=0$
$\Rightarrow\text{y}_2-4\text{y}_1+4\text{y}=0$
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$\frac{\text{dy}}{\text{dx}}=2\text{e}^{2\text{x}}\text{y}^2,\text{y}(0)=-1$