User:Egm6321.f11.team1.zheng.zx/personal place/HW2

= R*2.3 Show that the equation is linear in $${y}'$$ but an N1-ODE in general = Show that (2)p.7-6 is linear in y', and that (2) is in general an N1-ODE. But (2） is not the most general N1-ODE as represented by (1)p.7-6. Give an example of a more general N1-ODE.

Given
General N1-ODEs:
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$$  \displaystyle G({y}',y,x) =0 $$     (3-1) (1)p.7-6 Particular class of N1-ODEs: Linear in y'
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$$  \displaystyle M(x,y)+N(x,y){y}' =0 $$     (3-2) (2)p.7-6
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Find
1)Show that (2)p.7-6 is linear in y' 2)that (2) is in general an N1-ODE. 3)Give an example of a more general N1-ODE.

1) Show that (2)p.7-6 is linear in y'
For a operator to be linear, it has to satisfy the following:
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$$  \displaystyle F(\alpha u+\beta v)=\alpha F(u)+\beta F(v) ,\forall \alpha ,\beta \in \mathbb{R} $$     (3-3) Then we begin to check, Let Y = y',thus
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$$  \displaystyle G(x,y,Y)=\underbrace{M(x,y)}_{(1)}+\underbrace{N(x,y)Y}_{(2)} $$     (3-4) for (1)  We can regard this as a constant M; for (2)
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$$  \displaystyle F(x,y,Y)=N(x,y)Y $$     (3-5)
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$$  \displaystyle F(x,y,\alpha u+\beta v)=N(x,y)(\alpha u+\beta v)=\alpha N(x,y)u+\beta N(x,y)v $$     (3-6)
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$$  \displaystyle \alpha F(x,y,u)+\beta F(x,y,v)=\alpha N(x,y)u+\beta N(x,y)v $$     (3-7) So, Formula 3-6 is equal to 3-7, i.e.:
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$$  \displaystyle F(x,y,\alpha u+\beta v)=\alpha F(x,y,u)+\beta F(x,y,v) $$     (3-8) Then we can say that 3-2 i.e. (2)p.7-6 is linear in Y = y'
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2) Show that (2) is in general an N1-ODE
For a operator to be linear, it has to satisfy the following:
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$$  \displaystyle f(\alpha y_1+\beta y_2)=\alpha f(y_1)+\beta f(y_2) ,\forall \alpha ,\beta \in \mathbb{R} $$ Then we begin to check, So,
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$$  \displaystyle f(x,\alpha y_1+\beta y_2)\neq \alpha f(x,y_1)+\beta f(x,y_2) $$     (3-12) CONCLUSION: (2) is in general an N1-ODE
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3) Give an example of a more general N1-ODE
(2） is not the most general N1-ODE as represented by (1)p.7-6.
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$$  \displaystyle G(y',y,x)=0 $$     (3-13) A more general N1-ODE example can be:
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$$  \displaystyle M(x,y)+N(x,y)(y')^n=0 $$     (3-14) n is the power of y' This problem is solved by ourselves Contributed by Zexi
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= R*2.4 Verify that $$(4x^7+\sin y)+(x^2y^3)y'=0$$ is a N1-ODE =

Given

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$$  \displaystyle (4x^7+\sin y)+(x^2y^3)y'=0 $$     (4-1)
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Find
Formula 4-1 is a N1-ODE

Solution
1) The highest order of derivative is 1 and there is only one dependent variable in equation so it's a X1-ODE. (X can be L or N) 2) Verify Non-linearity For a operator to be linear, it has to satisfy the following:
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$$  \displaystyle F(\alpha u+\beta v)=\alpha F(u)+\beta F(v) ,\forall \alpha ,\beta \in \mathbb{R} $$     (4-2) Then we begin to check,
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$$  \displaystyle F(x,y)=(4x^7+\sin y)+(x^2y^3)y' $$     (4-3)
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$$  \displaystyle F(x,{\alpha}{y_1}+{\beta}{y_2})=4x^7 + sin({\alpha}{y_1}+{\beta}{y_2}) +{x^2}({\alpha}{y_1}+{\beta}{y_2})^3({\alpha}{y_1}+{\beta}{y_2})' $$     (4-4)
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$$  \displaystyle {\alpha}F(x,{y_1})+{\beta}F(x,{y_2})={\alpha}(4x^7+sin{y_1}+x^2{y_1}^3{y_1}')+{\beta}(4x^7+sin{y_2}+x^2{y_2}^3{y_2}') $$     (4-5)
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$$  \displaystyle F(x,{\alpha}{y_1}+{\beta}{y_2})\neq {\alpha}F(x,{y_1})+{\beta}F(x,{y_2}) $$     (4-6) So, it's nonlinear. CONCLUSION: The equation is a N1-ODE This problem is solved by ourselves Contributed by Zexi
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= R*2.10 Verify the solution for the N1-ODE =

Given

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$$  \displaystyle y(x)=\sin ^{-1}(k-15x^5) $$     (10-1) (1) p.10-4
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$$  \displaystyle \frac{d\phi (x,y(x))}{dx}=75x^4+(\cos y)y'=0 $$     (10-2) (2) p.8-6
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Find
Verify that 10-1 is the solution for the N1-ODE 10-2

Solution
We just need to put 10-1 into 10-2 and see if the equation is still in effect.
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$$  \displaystyle \frac{d\phi (x,y(x))}{dx}=75x^4+(\cos (\sin ^{-1}(k-15x^5)))(\sin^{-1}(k-15x^{5}))' $$     (10-3) We can easily verify that the equation 10-3 is equal to 0 with the help of Matlab. i.e.
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$$  \displaystyle 75x^4+(\cos (\sin ^{-1}(k-15x^5)))(\sin^{-1}(k-15x^{5}))'=0 $$     (10-4) Here are the codes:
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syms x syms k syms y y=asin(k-15*x^5); dy=diff(y,x); 75*x^4+cos(y)*dy

This problem is solved by ourselves Contributed by Zexi