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Question Number 169053 by Mastermind last updated on 23/Apr/22

Solve the ODE y′ + 2xy = xe^(−x^2 ) , with y(0)=1 Mastermind

$${Solve}\:{the}\:{ODE}\: \\ $$$${y}'\:+\:\mathrm{2}{xy}\:=\:{xe}^{−{x}^{\mathrm{2}} } ,\:{with}\:{y}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$$ \\ $$$${Mastermind} \\ $$

Answered by haladu last updated on 23/Apr/22

y′ + p(x)y = Q(x) p(x) = 2x I. f = e^(∫2x dx) = e^(x^2 ) y e^(x^2 ) = ∫ xe^(−x^2 ) × e^x^2 dx ye^x^2 = ∫ x dx = (x^2 /2) + C y = e^(−x^2 ) ( (x^2 /2) + C ) x = 0 y = 1 1 = C ⇒ y = e^(−x^2 ) ( (x^2 /2) + 1 ) Haladu B Bello

$$\:\:\boldsymbol{\mathrm{y}}'\:\:+\:\boldsymbol{\mathrm{p}}\left(\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{y}}\:=\:\boldsymbol{\mathrm{Q}}\left(\boldsymbol{\mathrm{x}}\right) \\ $$$$\:\:\:\boldsymbol{\mathrm{p}}\left(\boldsymbol{\mathrm{x}}\right)\:=\:\:\mathrm{2}\boldsymbol{\mathrm{x}}\:\: \\ $$$$\:\: \\ $$$$\:\:\boldsymbol{\mathrm{I}}.\:\boldsymbol{\mathrm{f}}\:\:=\:\:\boldsymbol{\mathrm{e}}^{\int\mathrm{2}\boldsymbol{\mathrm{x}}\:\:\boldsymbol{\mathrm{dx}}} \:\:=\:\:\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{x}}^{\mathrm{2}} \:} \:\: \\ $$$$\:\:\: \\ $$$$\:\:\:\boldsymbol{\mathrm{y}}\:\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{x}}^{\mathrm{2}} \:} \:\:=\:\:\:\int\:\:\boldsymbol{\mathrm{xe}}^{−\boldsymbol{\mathrm{x}}^{\mathrm{2}} } \:×\:\:\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{x}}^{\mathrm{2}} } \:\:\boldsymbol{\mathrm{dx}}\:\: \\ $$$$\:\:\:\: \\ $$$$\:\:\:\:\boldsymbol{\mathrm{ye}}^{\boldsymbol{\mathrm{x}}^{\mathrm{2}} } \:\:\:=\:\:\int\:\:\boldsymbol{\mathrm{x}}\:\:\boldsymbol{\mathrm{dx}}\:\:=\:\:\frac{\boldsymbol{\mathrm{x}}^{\mathrm{2}} }{\mathrm{2}}\:\:+\:\:\boldsymbol{\mathrm{C}} \\ $$$$\:\:\: \\ $$$$\:\:\:\:\:\:\boldsymbol{\mathrm{y}}\:\:=\:\:\boldsymbol{\mathrm{e}}^{−\boldsymbol{\mathrm{x}}^{\mathrm{2}} } \:\left(\:\:\frac{\boldsymbol{\mathrm{x}}^{\mathrm{2}} }{\mathrm{2}}\:\:+\:\:\boldsymbol{\mathrm{C}}\:\right) \\ $$$$\:\:\: \\ $$$$\:\:\:\boldsymbol{\mathrm{x}}\:\:=\:\:\mathrm{0}\:\:\:\:\boldsymbol{\mathrm{y}}\:\:\:=\:\:\mathrm{1}\:\: \\ $$$$\:\:\: \\ $$$$\:\:\:\mathrm{1}\:\:\:=\:\:\:\boldsymbol{\mathrm{C}} \\ $$$$\:\: \\ $$$$\:\:\Rightarrow\:\:\boldsymbol{\mathrm{y}}\:\:=\:\:\boldsymbol{\mathrm{e}}^{−\boldsymbol{\mathrm{x}}^{\mathrm{2}} } \:\left(\:\:\frac{\boldsymbol{\mathrm{x}}^{\mathrm{2}} }{\mathrm{2}}\:\:+\:\mathrm{1}\:\:\right) \\ $$$$\:\:\: \\ $$$$\:\:\boldsymbol{\mathrm{Haladu}}\:\:\boldsymbol{\mathrm{B}}\:\:\:\:\boldsymbol{\mathrm{Bello}} \\ $$

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