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.
Asked
Active
Viewed 238 times
3
-
2think of it this way, what function is its own derivative?? – The Integrator May 14 '18 at 13:07
-
1Picard–Lindelöf theorem. – Rodrigo de Azevedo May 14 '18 at 13:12
4 Answers
6
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 $$ z(t)e^{-t}=Const\implies z(t)=Ae^{t}. $$
Artem
- 14,414
3
Suppose $y$ is a solution to $y'=y$
Multiply both sides by $e^{-x}$ to get $$ y'e^{-x} = ye^{-x}$$
$$y'e^{-x}-ye^{-x}=0$$
$$ \frac {d}{dx} (ye^{-x}) =0$$ $$ye^{-x} =A$$ $$ y=Ae^{x} $$
Mohammad Riazi-Kermani
- 68,728
1
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.
Phil H
- 5,579
-
Thanks... but it's a little bit hard for me to understand. I should study more.. – 김기훈 May 15 '18 at 00:26
-
The question you were asking is essentially, will a different function yield the same areas between any 2 limits of integration? – Phil H May 15 '18 at 00:58
-
Really the opposite, is there another function with the same slope change as the integral? – Phil H May 15 '18 at 01:40
-1
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 $f(x) = A e^x$.
Jan
- 779