deborahambrozio
deborahambrozio deborahambrozio
  • 13-04-2017
  • Mathematics
contestada

Solve y' + 2xy = 2x^3, y(0) = 1

Respuesta :

LammettHash
LammettHash LammettHash
  • 13-04-2017
Multiply both sides by [tex]e^{x^2}[/tex], then you get

[tex]e^{x^2}+2xe^{x^2}y=2x^3e^{x^2}[/tex]
[tex]\left(e^{x^2}y\right)'=2x^3e^{x^2}[/tex]
[tex]e^{x^2}y=\displaystyle2\int x^3e^{x^2}\,\mathrm dx[/tex]
[tex]e^{x^2}y=e^{x^2}(x^2-1)+C[/tex]
[tex]y=x^2-1+Ce^{-x^2}[/tex]

Given that [tex]y(0)=1[/tex], you have

[tex]1=-1+Ce^0\implies C=2[/tex]

so that the particular solution is

[tex]y=x^2-1+2e^{-x^2}[/tex]
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