- #1
evinda
Gold Member
MHB
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Hello! (Wave)
I want to find the solution $\psi$ of the non-linear differential equation $y'=1+y^2$ that satisfies the condition $\psi(0)=0$. (Notice that the solution $\psi$ exists only for $- \frac{\pi}{2}< x < \frac{\pi}{2}$)
We notice that: $(tan^{-1})'(x)=\frac{1}{1+x^2} (\star) \left( -\frac{\pi}{2}<x< \frac{\pi}{2} \right)$.
So if $\psi$ is the solution of the initial value problem
$\left\{\begin{matrix}
y'=1+y^2\\ \\
y(0)=0
\end{matrix}\right.$
and thus $\psi$ exists at an interval that contains $0$, then for each $x$ that belongs to the interval $I$ of the solution $\psi$ we have:
$$\psi'(x)= 1+\psi^2(x) \ \ \forall x \in I$$
or equivalently $\frac{\psi'(x)}{1+ \psi^2(x)}=1$.
From the relation $(\star)$ we have $(tan^{-1} \psi(t))'=1 \ \ \forall t \in I$.
We integrate the last relation from $0$ to $x$ ($ \ \forall x \in I$) and we have:$$tan^{-1} \psi(x)- \tan^{-1} \psi(0)=x \ \ \forall x \in I \Rightarrow \psi(x)= \tan x \ \ \forall x \in I$$
$\psi(x)$ has as maximal existence interval the interval $\left( - \frac{\pi}{2}, \frac{\pi}{2}\right)$ and we can easily verify that $\psi(x)= \tan x$ satisfies the initial value problem $\left\{\begin{matrix}
y'=1+y^2\\ \\
y(0)=0
\end{matrix}\right.$.Firstly, having found the solution $\psi(x)= \tan x$ do we have to verify it? (Thinking)Also with existence interval is the domain of the function meant? If so, isn't it $\mathbb{R}$ ? Or am I wrong? (Thinking)
I want to find the solution $\psi$ of the non-linear differential equation $y'=1+y^2$ that satisfies the condition $\psi(0)=0$. (Notice that the solution $\psi$ exists only for $- \frac{\pi}{2}< x < \frac{\pi}{2}$)
We notice that: $(tan^{-1})'(x)=\frac{1}{1+x^2} (\star) \left( -\frac{\pi}{2}<x< \frac{\pi}{2} \right)$.
So if $\psi$ is the solution of the initial value problem
$\left\{\begin{matrix}
y'=1+y^2\\ \\
y(0)=0
\end{matrix}\right.$
and thus $\psi$ exists at an interval that contains $0$, then for each $x$ that belongs to the interval $I$ of the solution $\psi$ we have:
$$\psi'(x)= 1+\psi^2(x) \ \ \forall x \in I$$
or equivalently $\frac{\psi'(x)}{1+ \psi^2(x)}=1$.
From the relation $(\star)$ we have $(tan^{-1} \psi(t))'=1 \ \ \forall t \in I$.
We integrate the last relation from $0$ to $x$ ($ \ \forall x \in I$) and we have:$$tan^{-1} \psi(x)- \tan^{-1} \psi(0)=x \ \ \forall x \in I \Rightarrow \psi(x)= \tan x \ \ \forall x \in I$$
$\psi(x)$ has as maximal existence interval the interval $\left( - \frac{\pi}{2}, \frac{\pi}{2}\right)$ and we can easily verify that $\psi(x)= \tan x$ satisfies the initial value problem $\left\{\begin{matrix}
y'=1+y^2\\ \\
y(0)=0
\end{matrix}\right.$.Firstly, having found the solution $\psi(x)= \tan x$ do we have to verify it? (Thinking)Also with existence interval is the domain of the function meant? If so, isn't it $\mathbb{R}$ ? Or am I wrong? (Thinking)
