Question

# If y = ex (sin x + cos x) prove that $\frac{{d}^{2}y}{d{x}^{2}}-2\frac{dy}{dx}+2y=0$.

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Solution

## Here, $y={e}^{x}\left(\mathrm{sin}x+\mathrm{cos}x\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.x\mathit{,}\mathit{}\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}={e}^{x}\left(\mathrm{sin}x+\mathrm{cos}x\right)+{e}^{x}\left(\mathrm{cos}x-\mathrm{sin}x\right)=2{e}^{x}\mathrm{cos}x\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.x\mathit{,}\mathit{}\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}\frac{{\mathit{d}}^{\mathit{2}}\mathit{y}}{\mathit{d}{\mathit{x}}^{\mathit{2}}}=2{e}^{x}\mathrm{cos}x-2{e}^{x}\mathrm{sin}x\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}\mathrm{LHS}=\frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^{\mathit{2}}}-2\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}+2y\phantom{\rule{0ex}{0ex}}=2{e}^{x}\mathrm{cos}x-2{e}^{x}\mathrm{sin}x-4{e}^{x}\mathrm{cos}x+2{e}^{x}\left(\mathrm{sin}x+\mathrm{cos}x\right)\phantom{\rule{0ex}{0ex}}=0=\mathrm{RHS}\phantom{\rule{0ex}{0ex}}$ Hence proved.

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