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#### karush

##### Well-known member

- Jan 31, 2012

- 2,886

find the general solution for

$y' + 2y = 2 - {{\bf{e}}^{ - 4t}} \hspace{0.25in}y\left( 0 \right) = 1$

find

$\displaystyle u(x)=\exp \left(\int 2 \, dt\right) =e^{2t}$

$e^{2t}y' + 2\frac{e^{2t}}{2}y =2 e^{2t} - {{\bf{e}}^{ - 4t}}e^{2t}$

$(e^{2t}y)'=2e^{2t}-e^{-2t}$

i continued but didn't get the answer which is

$y\left( t \right) = 1 + \frac{1}{2}{{\bf{e}}^{ - 4t}} - \frac{1}{2}{{\bf{e}}^{ - 2t}}$

$y' + 2y = 2 - {{\bf{e}}^{ - 4t}} \hspace{0.25in}y\left( 0 \right) = 1$

find

$\displaystyle u(x)=\exp \left(\int 2 \, dt\right) =e^{2t}$

$e^{2t}y' + 2\frac{e^{2t}}{2}y =2 e^{2t} - {{\bf{e}}^{ - 4t}}e^{2t}$

$(e^{2t}y)'=2e^{2t}-e^{-2t}$

i continued but didn't get the answer which is

$y\left( t \right) = 1 + \frac{1}{2}{{\bf{e}}^{ - 4t}} - \frac{1}{2}{{\bf{e}}^{ - 2t}}$

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