Respuesta :

LammettHash
[tex]y'-y=11te^{2t}[/tex]
[tex]e^{-t}y'-e^{-t}y=11te^t[/tex]
[tex]\dfrac{\mathrm d}{\mathrm dt}\left[e^{-t}y\right]=11te^t[/tex]
[tex]e^{-t}y=11\displaystyle\int te^t\,\mathrm dt[/tex]
[tex]e^{-t}y=11e^t(t-1)+C[/tex]

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

[tex]1=11(-1)+C\implies C=12[/tex]

and so

[tex]e^{-t}y=11e^t(t-1)+12[/tex]
[tex]y=11e^{2t}(t-2)+12e^t[/tex]

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