- #1

karush

Gold Member

MHB

- 3,269

- 5

$\tiny{b.1.2.2a}$

$\dfrac{dy}{dt}=y-5, \quad y(0)=y_0$$\begin{array}{rll}\displaystyle

y'&=y-5\\

y'-y&=-5\\

u(x)&=\exp\int-1\, dx&= e^{-t}\\

(e^{-t}y)'&=-5e^{-t}\\

e^{-t}y&=-5\int e^{-t} dt&= -5e^{-t}+c\\

y&=-5\dfrac{e^{-t}}{e^{-t}}+\dfrac{c}{e^{-t}}&=-5e+ce^t\\

y(0)&=5+C=y_0\\

\implies C&=y_0-5\\

y&=5+(y_0-5)e^t

\end{array}$

ok I think this is correct... wasn't sure about the 5 scalar

typo's maybe :unsure:

$\dfrac{dy}{dt}=y-5, \quad y(0)=y_0$$\begin{array}{rll}\displaystyle

y'&=y-5\\

y'-y&=-5\\

u(x)&=\exp\int-1\, dx&= e^{-t}\\

(e^{-t}y)'&=-5e^{-t}\\

e^{-t}y&=-5\int e^{-t} dt&= -5e^{-t}+c\\

y&=-5\dfrac{e^{-t}}{e^{-t}}+\dfrac{c}{e^{-t}}&=-5e+ce^t\\

y(0)&=5+C=y_0\\

\implies C&=y_0-5\\

y&=5+(y_0-5)e^t

\end{array}$

ok I think this is correct... wasn't sure about the 5 scalar

typo's maybe :unsure:

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