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dy-dx-2xy-x-2-y-0-3-Solve-

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Question Number 181618 by Mastermind last updated on 27/Nov/22
(dy/dx)+2xy=x^2 y(0)=3 Solve .
$$\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{2xy}=\mathrm{x}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{y}\left(\mathrm{0}\right)=\mathrm{3} \\ $$$$ \\ $$$$\mathrm{Solve} \\ $$$$ \\ $$$$. \\ $$
Answered by hmr last updated on 27/Nov/22
multiply both sides of eq. to a function like μ(x). μ(x) (dy/dx) + 2μ(x) xy = μ(x) x^2 try to form the LHS like this: (d/dx)(μ(x) y) = (dμ/dx) y + μ(x) (dy/dx) so blue ones should be equal. (dμ/dx) y = 2μ(x) xy ((dμ/dx)/(μ(x))) = 2x (d/dx)(ln∣μ(x)∣) = 2x ln∣μ(x)∣ = x^2 + c μ (x) = ±e^c e^x^2 = C e^x^2 C = 1 → μ(x) = e^x^2 replace μ(x) in equation. (d/dx)(e^x^2 y) = e^x^2 x^2 e^x^2 y = ∫e^x^2 x^2 dx y = ((∫e^x^2 x^2 dx + c)/e^x^2 )
$${multiply}\:{both}\:{sides}\:{of}\:{eq}. \\ $$$${to}\:{a}\:{function}\:{like}\:\mu\left({x}\right). \\ $$$$\mu\left({x}\right)\:\frac{{dy}}{{dx}}\:+\:\mathrm{2}\mu\left({x}\right)\:{xy}\:=\:\mu\left({x}\right)\:{x}^{\mathrm{2}} \\ $$$$ \\ $$$${try}\:{to}\:{form}\:{the}\:{LHS}\:{like}\:{this}: \\ $$$$\frac{{d}}{{dx}}\left(\mu\left({x}\right)\:{y}\right)\:=\:\frac{{d}\mu}{{dx}}\:{y}\:+\:\mu\left({x}\right)\:\frac{{dy}}{{dx}}\: \\ $$$$ \\ $$$${so}\:{blue}\:{ones}\:{should}\:{be}\:{equal}. \\ $$$$\frac{{d}\mu}{{dx}}\:{y}\:=\:\mathrm{2}\mu\left({x}\right)\:{xy}\: \\ $$$$\frac{{d}\mu/{dx}}{\mu\left({x}\right)}\:=\:\mathrm{2}{x} \\ $$$$\frac{{d}}{{dx}}\left({ln}\mid\mu\left({x}\right)\mid\right)\:=\:\mathrm{2}{x} \\ $$$${ln}\mid\mu\left({x}\right)\mid\:=\:{x}^{\mathrm{2}} \:+\:{c} \\ $$$$\mu\:\left({x}\right)\:=\:\pm{e}^{{c}} \:{e}^{{x}^{\mathrm{2}} } \:=\:{C}\:{e}^{{x}^{\mathrm{2}} } \\ $$$${C}\:=\:\mathrm{1}\:\rightarrow\:\mu\left({x}\right)\:=\:{e}^{{x}^{\mathrm{2}} } \\ $$$$ \\ $$$${replace}\:\mu\left({x}\right)\:{in}\:{equation}. \\ $$$$\frac{{d}}{{dx}}\left({e}^{{x}^{\mathrm{2}} } \:{y}\right)\:=\:{e}^{{x}^{\mathrm{2}} } \:{x}^{\mathrm{2}} \\ $$$${e}^{{x}^{\mathrm{2}} } \:{y}\:=\:\int{e}^{{x}^{\mathrm{2}} } \:{x}^{\mathrm{2}} \:{dx} \\ $$$${y}\:=\:\frac{\int{e}^{{x}^{\mathrm{2}} } \:{x}^{\mathrm{2}} \:{dx}\:+\:{c}}{{e}^{{x}^{\mathrm{2}} } }\: \\ $$$$ \\ $$
Answered by ali009 last updated on 28/Nov/22
using the rule of linear first order eq (dy/dx)+p(x)y=Q(x) y=(1/(D(x)))∫D(x)Q(x)dx D(x)=e^(∫p(x)dx) =e^(∫2xdx) =e^x^2 y=(1/e^x^2 )(∫e^x^2 x^2 dx+c)
$${using}\:{the}\:{rule}\:{of}\:{linear}\:{first}\:{order}\:{eq} \\ $$$$\frac{{dy}}{{dx}}+{p}\left({x}\right){y}={Q}\left({x}\right) \\ $$$${y}=\frac{\mathrm{1}}{{D}\left({x}\right)}\int{D}\left({x}\right){Q}\left({x}\right){dx} \\ $$$${D}\left({x}\right)={e}^{\int{p}\left({x}\right){dx}} ={e}^{\int\mathrm{2}{xdx}} ={e}^{{x}^{\mathrm{2}} } \\ $$$${y}=\frac{\mathrm{1}}{{e}^{{x}^{\mathrm{2}} } }\left(\int{e}^{{x}^{\mathrm{2}} } {x}^{\mathrm{2}} {dx}+{c}\right) \\ $$

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