User:Egm6321.f11.team2.Xia/RP2.3

=Problem 2.3- Linear ODE=

Given
An ODE of 1st order is read as

$$\displaystyle M(x,y)\,dx+N(x,y)\,y'=0 $$ (2.03.1)

Find
show that the eqn (2.03.1) is linear in $$\displaystyle y'$$, and that it is in general an N1-ODE. But it is not the most general N1-ODE as represented by $$ \displaystyle G(y',y,x)=0 $$ (2.03.2) Give an example of a more general N1-ODE.

Definition of Linear
If $$\displaystyle L_{1} $$ is a linear operator, then it must satisfy $$\displaystyle L_{1}(\alpha u+\beta v)=\alpha L_{2}(u)+\beta L_{2}(v)$$ Where $$\displaystyle u,v $$ are the independent variables, and $$\displaystyle \alpha, \beta \in \mathbb{R}.$$ $$\displaystyle L_{1}(\alpha y'_{1}+\beta y'_{2})=-[\alpha(\frac{M}{N})_{1}+\beta(\frac{M}{N})_{2}]$$ $$\displaystyle L_{1}(\alpha y'_{1}+\beta y'_{2})=-\alpha(\frac{M}{N})_{1}-\beta(\frac{M}{N})_{2}$$ $$\displaystyle L_{1}(\alpha y'_{1}+\beta y'_{2})=\alpha L_{1}(y'i_{1})+\beta L_{1}(y'_{2})$$

$$\displaystyle \therefore M(x,y)\,dx+N(x,y)\,y'=0 $$ is  linear in $$\displaystyle y' $$.

And it is easy to proof that $$\displaystyle L_{1}(\alpha y_{1}+\beta y_{2})\neq\alpha L_{1}(y_1)+\beta L_{1}(y_2)$$ So, is in general an N1-ODE.

An example of the general N1-ODE
$$ \displaystyle M(x,y)+N(x,y)f(y')=0 $$. (2.03.3) where $$ \displaystyle f(y') $$ is an arbitrary function of $$ \displaystyle y' $$.