Solução da equação diferencial linear: t.y'(t) - 2.y(t) = - t
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1. A su forma estándar:
![y'-\dfrac{2}{t}y=-1 y'-\dfrac{2}{t}y=-1](https://tex.z-dn.net/?f=y%27-%5Cdfrac%7B2%7D%7Bt%7Dy%3D-1)
restricción:![t\neq 0 t\neq 0](https://tex.z-dn.net/?f=t%5Cneq+0)
2. Factor integrante:
![\displaystyle
F.I.=\exp\int -\dfrac{2}{t}dt=\exp(-2\ln|t|)=\dfrac{1}{t^2} \displaystyle
F.I.=\exp\int -\dfrac{2}{t}dt=\exp(-2\ln|t|)=\dfrac{1}{t^2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%0AF.I.%3D%5Cexp%5Cint+-%5Cdfrac%7B2%7D%7Bt%7Ddt%3D%5Cexp%28-2%5Cln%7Ct%7C%29%3D%5Cdfrac%7B1%7D%7Bt%5E2%7D)
3. multiplicando a la ecuación en (1) por el factor integrante:
![\dfrac{1}{t^2}y'-\dfrac{2}{t^3}y=-\dfrac{1}{t^2} \dfrac{1}{t^2}y'-\dfrac{2}{t^3}y=-\dfrac{1}{t^2}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7Bt%5E2%7Dy%27-%5Cdfrac%7B2%7D%7Bt%5E3%7Dy%3D-%5Cdfrac%7B1%7D%7Bt%5E2%7D)
4. resolviendo
![\displaystyle
\dfrac{1}{t^2}y'-\dfrac{2}{t^3}y=-\dfrac{1}{t^2}\\ \\
\left(\dfrac{1}{t^2}y\right)'=-\dfrac{1}{t^2}\\ \\
\dfrac{1}{t^2}y=\int-\dfrac{1}{t^2}dt\\ \\
\dfrac{1}{t^2}y=\dfrac{1}{t} +C\\ \\
\boxed{y=t+Ct^2}
\displaystyle
\dfrac{1}{t^2}y'-\dfrac{2}{t^3}y=-\dfrac{1}{t^2}\\ \\
\left(\dfrac{1}{t^2}y\right)'=-\dfrac{1}{t^2}\\ \\
\dfrac{1}{t^2}y=\int-\dfrac{1}{t^2}dt\\ \\
\dfrac{1}{t^2}y=\dfrac{1}{t} +C\\ \\
\boxed{y=t+Ct^2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%0A%5Cdfrac%7B1%7D%7Bt%5E2%7Dy%27-%5Cdfrac%7B2%7D%7Bt%5E3%7Dy%3D-%5Cdfrac%7B1%7D%7Bt%5E2%7D%5C%5C+%5C%5C%0A%5Cleft%28%5Cdfrac%7B1%7D%7Bt%5E2%7Dy%5Cright%29%27%3D-%5Cdfrac%7B1%7D%7Bt%5E2%7D%5C%5C+%5C%5C%0A%5Cdfrac%7B1%7D%7Bt%5E2%7Dy%3D%5Cint-%5Cdfrac%7B1%7D%7Bt%5E2%7Ddt%5C%5C+%5C%5C%0A%5Cdfrac%7B1%7D%7Bt%5E2%7Dy%3D%5Cdfrac%7B1%7D%7Bt%7D+%2BC%5C%5C+%5C%5C%0A%5Cboxed%7By%3Dt%2BCt%5E2%7D%0A)
restricción:
2. Factor integrante:
3. multiplicando a la ecuación en (1) por el factor integrante:
4. resolviendo
AninhaBR02:
Agradeço! Mas parece que o formato n é suportado rs
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