Possible solutions to different types of differential equations Here are some EE exam questions. Very similar questions.


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*What are two necessary conditions that need to be fulfilled so there's only one solution to: $y'=f(x,y)$ which satisfies $y(x_0)=y_0$ (initial condition)and $f:D \to R$,   $D⊆R^2$, $(x_0,y_0)∈D$


-$f$ and $ df/dy$ are continous. Any other?


*Prove non-existence of 3rd order differential equation $y'''=f(x,y,y',y'')$  that fulfills theorem of existence and uniqueness of solutions and which 2 solutions are $\varphi_1(x)=x,  \varphi_2(x)=\sin x$.


-if it was a homogeneous linear with constant coefficients, then I would know there are 4 solutions and you need 4th order equation. What am I missing from the theorem?
 A: About question 1.
I have several comments here:


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*It is strange for me to request necessary conditions. I would rather ask for sufficient conditions. In particular if you have in mind to apply Picard–Lindelöf theorem. For example the IVP problem $$y^\prime=1+y^{\frac{2}{3}}, y(0)=0$$ is such that $\frac{\partial f}{\partial y}$ doesn't exist at the origin. However it has a unique local solution.

*Continuity of $f$ and of $\frac{\partial f}{\partial y}$ is sufficient to apply Picard–Lindelöf theorem...

*But that gives only the existence and unicity of a local solution! Which means that the solution might not exist in all $D$.


About question 2.
Suppose that $$y^{\prime \prime \prime}=f(x,y,y^\prime,y^{\prime \prime})$$ is a 3rd order differential equation satisfying for example the Picard–Lindelöf theorem. Then you have a unique solution $y(x)$ (at least in the neighborhood of $0$) such that $$y(0)=y_0, y^\prime(0)=y_1,y^{\prime \prime}(0)=y_2.$$ However, we have $$\varphi_1(0)=\varphi_2(0)=0, \varphi_1^\prime(0)=\varphi_2^\prime(0)=1, \varphi_1^{\prime \prime}(0)=\varphi_2^{\prime \prime}(0)=0$$ but $\varphi_1 \neq \varphi_2$ for any neighborhood of $0$. So $\varphi_1, \varphi_2$ cannot be two solutions of our 3rd order differential equation.
