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$\dfrac{dY}{dx}=Y$, where $Y=\begin{pmatrix}y_1(x) \\ y_2(x)\end{pmatrix}$. Also, $Y(0)=\begin{pmatrix} 0\\ 1\end{pmatrix}$.

Then both solutions $y_1(x)$ and $y_2(x)$ are

  1. increasing functions
  2. decreasing functions
  3. neither increasing nor decreasing
  4. constant functions.

After solving, I got $y_1(x)=A \exp(x)$, $y_2(x)=B \exp(x)$. Boundary conditions give $y_1(x)= 0$ and $y_2(x)=\exp(x)$. The 4th option is incorrect but can not select any option out of remaining three. Please help me.

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    $y_1(0)=0$ is not satisfied by your solution. Also, does $\exp(x)$ ever decreases ?2012-12-24
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    @WhitAngl: how does $y_1(x) = 0$ not satisfy $y_1(0) = 0$?2012-12-24
  • 1
    There may be a clash of language here: in English it's common these days to use "increasing" to mean strictly increasing ($x_1 < x_2$ implies $f(x_1) < f(x_2)$), while "nondecreasing" means $x_1 < x_2$ implies $f(x_1) \le f(x_2)$. Similarly for "decreasing" and "nonincreasing" with $f(x_1) > f(x_2)$ and $f(x_1) \ge f(x_2)$ respectively. However, some people consider "increasing" and "decreasing" to mean "nondecreasing" and "nonincreasing". In this case both $y_1$ and $y_2$ are nondecreasing, but only one is strictly increasing.2012-12-24
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    y1(0) = 0 gives A=0, so y1(x)=0< then how is it not satisfied?2012-12-24

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