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Consider the Initial Value Problem

$\dfrac{dy}{dx}=\dfrac y{x-1}$, $y(a)=b$

i) explain why the IVP has a unique solution if $a$ is not equal to $1$.

ii) explain why the IVP has infinitely many solutions if $a=1$, $b=0$

Would someone be able to help/explain these to me? Thanks in advance!

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write your equation in the form $$\frac{dy}{y}=\frac{dx}{x-1}$$ and integrate

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  • $\begingroup$ Thanks, helped a lot! $\endgroup$ – RBadger Mar 4 '18 at 21:44
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We have $$\dfrac{dy}{dx}=\dfrac y{x-1}\implies \int\frac1y\,dy=\int\frac1{x-1}\,dx\implies \ln y=\ln(x-1)+C$$ where $C$ is a constant. Taking exponentials, $$y=e^C\cdot(x-1)$$ which is unique when $a\neq1$ as $C$ is fixed.

But when $a=1,b=0$, we have $$0=e^C\cdot0$$ so $C$ can take infinitely many values.

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  • $\begingroup$ Haha yeah I got it in the end, thanks a lot :) $\endgroup$ – RBadger Mar 4 '18 at 21:45
  • $\begingroup$ You´ve forgotten to mark the answer as accepted. Check your other questions as well, please. $\endgroup$ – callculus Apr 24 at 16:05

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