Why ODE Linear? Initial Value Problem Explanation

[evinda]

Hello! (Wave)

I am looking at initial value problems for ordinary differential equations.Let $a,b, \ a<b, \ f: [a,b] \times \mathbb{R} \to \mathbb{R}$ function and $y_0 \in \mathbb{R}$.We are looking for a $y: [a,b] \to \mathbb{R}$ such that$$(1)\left\{\begin{matrix}
y'(t)=f(t,y(t))\\
y(a)=y_0
\end{matrix}\right.$$Each function $y \in C^{1}[a,b]$ that satisfies the differential equation of $(1)$ and the initial value $y(a)=y_0$ is called solution of the initial value problem $(1)$.Special caseLet $f$ be a polynomial of degree $1$ as for $y$. Then the corresponding ODE is called linear, and the problem $(1)$ is written as:$$(2)\left\{\begin{matrix}
y'(t)=p(t)y(t)+q(t), a \leq t \leq b\\
y(a)=y_0
\end{matrix}\right.$$It holds that $deg_y q(t)< deg_y y(t)=1 \Rightarrow q(y)=\text{ constant }$, right?

But, if so then why is in this case the ODE linear? (Thinking)

 

[John Harris1]

...
It holds that $deg_y q(t)< deg_y y(t)=1 \Rightarrow q(y)=\text{ constant }$, right?

But, if so then why is in this case the ODE linear? (Thinking)

No. It is true that
deg_y q(t)< deg_y y(t)
But if
q(t) = e^t2
the deg_y q(t) = 0 which is less than 1.

Also the linearity of the ODE has nothing to do with q. If the exponent for y and all of the derivative in the ODE [y, y', y'', y''', ...] are 1, then the ODE is called linear. That is, it is linear in y and its derivatives.

 

[evinda]

No. It is true that
deg_y q(t)< deg_y y(t)
But if
q(t) = e^t2
the deg_y q(t) = 0 which is less than 1.

Also the linearity of the ODE has nothing to do with q. If the exponent for y and all of the derivative in the ODE [y, y', y'', y''', ...] are 1, then the ODE is called linear. That is, it is linear in y and its derivatives.

Nice! Thank you! (Smile)