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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 ? –  WhitAngl Dec 24 '12 at 6:11
    
@WhitAngl: how does $y_1(x) = 0$ not satisfy $y_1(0) = 0$? –  Robert Israel Dec 24 '12 at 6:18
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. –  Robert Israel Dec 24 '12 at 6:26
    
y1(0) = 0 gives A=0, so y1(x)=0< then how is it not satisfied? –  Alka Goyal Dec 24 '12 at 6:33
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1 Answer

Note that for solutions of $y_1,y_2$ you can also have constants involved; the general solutions are (without taking initial conditions into consideration) $y_1=a_1e^x+c_1$ and $y_2=a_2e^x+c_2$.

However, differentiating, we get $y_1'=a_1e^x=0\Rightarrow a_1=0$ and $y_2'=a_2e^x=1\Rightarrow a_2=1$. Substituting these constants back in, we see $$y_1(x)=c$$ and $$y_2(x)=e^x$$

Therefore, $y_1$ is constant (and therefore neither increasing nor decreasing), and $y_2$ is increasing (neither constant nor decreasing). Therefore none of $(1)-(4)$ are satisfied.

Note, that, however, if we say $f$ is increasing iff $x_1>x_2$ implies only that $f(x_1)\geq f_2(x_2)$, thereby including constant functions as increasing, we have $(1)$: both functions $y_1,y_2$ satisfy this condition.

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