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In school, I have recently been learning about simple differential equations. We know that the solution of $y'=y$ is $y=Ae^x$, where $A$ is a constant. But how can we know that it is the only solution? The only thing I can figure out is that $y$ is continuously differentiable. Help me, please.

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marked as duplicate by Hans Lundmark, Dando18, Strants, Matthew Leingang, B. Mehta May 14 '18 at 17:20

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Let $z$ be a solution to $y'=y$. Consider $z(t)e^{-t}$. We have $$ \frac{d}{dt}(z(t)e^{-t})=z'(t)e^{-t}-z(t)e^{-t}=0, $$ therefore $$ z(t)e^{-t}=Const\implies z(t)=Ae^{t}. $$

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Suppose $y$ is a solution to $y'=y$

Multiply both sides by $e^{-x}$ to get $$ y'e^{-x} = ye^{-x}$$

$$y'e^{-x}-ye^{-x}=0$$

$$ \frac {d}{dx} (ye^{-x}) =0$$ $$ye^{-x} =A$$ $$ y=Ae^{x} $$

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I've heard this kind of question before. An anti-derivative will yield a definitive delta area under a graph of a function between any 2 limits. Seeing there is only one delta area, any different expressions defining it would essentially be the same.

The area of a right triangle $1/2xy$ or $1/2x^2 \tan \theta$ are the same with a trigonometric substitution. For the family of anti-derivatives whose only difference is C, where C makes no difference in calculating the definite integral, in reverse they only have one derivative.

To summarize, a function that defines areas of different regions under a graph is unique and so those different areas are themselves defined by a unique function under which they exist.

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  • $\begingroup$ Thanks... but it's a little bit hard for me to understand. I should study more.. $\endgroup$ – 김기훈 May 15 '18 at 0:26
  • $\begingroup$ The question you were asking is essentially, will a different function yield the same areas between any 2 limits of integration? $\endgroup$ – Phil H May 15 '18 at 0:58
  • $\begingroup$ Really the opposite, is there another function with the same slope change as the integral? $\endgroup$ – Phil H May 15 '18 at 1:40
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Suppose there is another solution $y=f(x)$. Then we must have that $y=f(x)-Ae^x$ is a solution. Now working this out gives us: $$y'=\frac{\partial }{\partial x}(f(x) - A e^x) = f'(x) - A e^x$$ Note that we must have $y=y'$ so $f(x)=f'(x)$ clearly this only holds for $f(x) = A e^x$.

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    $\begingroup$ This is a circular argument. $\endgroup$ – JP McCarthy May 14 '18 at 13:17

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