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Question Number 170410 by Mastermind last updated on 23/May/22

$$Giveny={xe}^{-x},showthat$$$$\frac{d^{2}y}{{dx}^2}+2\frac{dy}{dx}+y=0$$$$$$$$Mastermind$$

Answered by floor(10²Eta[1]) last updated on 23/May/22

$$\frac{\mathrm{dy}}{\mathrm{dx}}={\mathrm{e}}^{-\mathrm{x}}-{\mathrm{xe}}^{-\mathrm{x}}$$$$\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=-{2\mathrm{e}}^{-\mathrm{x}}+{\mathrm{xe}}^{-\mathrm{x}}$$$$-{2\mathrm{e}}^{-\mathrm{x}}+{\mathrm{xe}}^{-\mathrm{x}}+2({\mathrm{e}}^{-\mathrm{x}}-{\mathrm{xe}}^{-\mathrm{x}})+{\mathrm{xe}}^{-\mathrm{x}}=0$$$$$$

Commented by Mastermind last updated on 24/May/22

$$Thanksforansweringmyquestion$$

Answered by MikeH last updated on 23/May/22

$$\frac{dy}{dx}=-{xe}^{-x}+e^{-x}$$$$\frac{d^{2}y}{{dx}^2}={xe}^{-x}-2e^{-x}$$$$\frac{d^{2}y}{{dx}^2}+2\frac{dy}{dx}+y={xe}^{-x}-2e^{-x}-2{xe}^{-x}+2e^{-x}+{xe}^{-x}=0$$

Commented by Mastermind last updated on 24/May/22

$$$$$$Thanksman$$

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