Please guide me on this one. We have to choose one of the $4$ options.

Let $y_1(x)$ and $y_2(x)$ be the solutions of the differential equation $\frac{dy}{dx}=y+17$ with initial conditions $y_1(0)=0$, $y_2(0)=1$. Then

$1.$ $y_1$ and $y_2$ will never intersect.

$2.$ $y_1$ and $y_2$ will intersect at $x=17$.

$3.$ $y_1$ and $y_2$ will intersect at $x=e$.

$4.$ $y_1$ and $y_2$ will intersect at $x=e$.

Now, $(\frac{1}{y+17})dy=dx,$ Integrating, $ln|y+17|=x+c$

If I put, $x=0$ and $y=0$, I get $ln17=c$. If I put $x=0$ and $y=1$, I get $ln18=c$. Can that be possible? I wish somebody could show me the right path.

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    $\begingroup$ When you integrated, why did you take the natural log of both sides? You should have gotten $\ln|y+17| = x + C$. Then exponentiate from there and solve for $y_{1}$ and $y_{2}$ given the initial values. $\endgroup$ – Alex Wertheim Aug 12 '13 at 16:12
  • $\begingroup$ oh yes, thanks! $\endgroup$ – Ramit Aug 12 '13 at 16:13
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    $\begingroup$ No problem! Feel free to comment again if you need more help. :) $\endgroup$ – Alex Wertheim Aug 12 '13 at 16:13
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    $\begingroup$ Well, for different functions, you get different integrating constants. But you're almost there: $y_1(x)=17e^x-17$ and $y_2(x)=18e^x-17$ the rest should be easy. $\endgroup$ – walcher Aug 12 '13 at 16:23
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    $\begingroup$ @Ramit, see walcher's comment. You are plugging in two different initial conditions. That you get two different values is entirely fine and consistent. Perhaps label one constant $C_{1}$ and one constant $C_{2}$. $\endgroup$ – Alex Wertheim Aug 12 '13 at 16:24

We get

$y_1=17e^x -17$ and

$y_2=18e^x -17$.

Hence, the pair of equations does not intersect at any point.

If we equate $y_1$ and $y_2$, we get $e^x=0$ which is not defined.


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