User:Egm6321.f10.team2.dube/HW2

=Problem 1: Verify Mdx+Ndy=0 is general N1_ODE=

Given
In general form a N1_ODE can be written as follows:


 * {| style="width:100%" border="0"

$$M\left ( x,y \right )dx+N\left ( x,y \right )dy=0 \displaystyle$$ (1.1)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

Find
Show Eq1 has a general form.

Solution
We can write Eq as


 * {| style="width:100%" border="0"

$$M\left ( x,y \right )+N\left ( x,y \right )\frac{dy}{dx}=M\left ( x,y \right )+N\left ( x,y \right )y^{'}\displaystyle$$ (1.2)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

Now let consider about general form as follows :


 * {| style="width:100%" border="0"

$$P\left ( x \right )y{}'+Q\left ( x \right )y=R\left ( x \right )\displaystyle$$ (1.3)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

You can see from the lecturer notes from the 10-2

If we divide both sides with P(x);


 * {| style="width:100%" border="0"

$$y{}'+\frac{Q(x)}{P(x)}y=\frac{R(x)}{P(x)}\displaystyle$$ (1.4)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

Then we can say that ;
 * {| style="width:100%" border="0"

$$\bar{N}(x,y)=1\displaystyle$$ (1.5)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }


 * {| style="width:100%" border="0"

$$\bar{M}(x,y)=\frac{Q(x)}{P(x)}y-\frac{R(x)}{P(x)}\displaystyle$$ (1.6) So we concluded that
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }


 * {| style="width:100%" border="0"

$$\bar{M}(x,y)+\bar{N}(x,y)y{}'=0\displaystyle$$ (1.7)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

Another example for this porblem we can give in book that it says in the page 511 for the Variables Separable kind of differential equations.

It is shown as :


 * {| style="width:100%" border="0"

$$g(y)\frac{dy}{dt}=f(t)\displaystyle$$ (1.8)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

To solve this ;


 * {| style="width:100%" border="0"

$$\int g(y)\frac{dy}{dt}=\int f(t)\displaystyle$$ (1.9)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }


 * {| style="width:100%" border="0"

$$\int g(y)dy=\int f(t)dt\displaystyle$$ (1.10)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

We can rearrange this


 * {| style="width:100%" border="0"

$$g(y)\frac{dy}{dt}-f\left ( t \right )=0\displaystyle$$ (1.11)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

And again


 * {| style="width:100%" border="0"

$$\bar{N}(t,y)=g(y),\bar{M}(t,y)=-f(t)\displaystyle$$ (1.12)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }


 * {| style="width:100%" border="0"

$$\bar{N}(t,y)\frac{dy}{dt}+\bar{M}(t,y)=0\displaystyle$$ (1.13)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

By showing this now we can say that for every type first order nonlinear differansial equations can be written in this type by some arrangements.

=Problem 2 : Show why Eq 7 in page (7-1) is first order and non linear=

Given
Eq 7 in page 7-1
 * {| style="width:100%" border="0"

$$\left ( 4*x^7+sin(y) \right )+\left ( x^2*y^3 \right )\frac{\mathrm{d} y}{\mathrm{d} x}=0 \displaystyle$$ (2.1)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

Find
Show why the equation above is first order and non linear

Solution
Since we define our first order differential $$F\left ( x,y,y{}' \right )=0$$ there must be only first degree differential of dependent variable in the function.If we look at the example we can notice that there is only first degree differential of dependent variable multiplied by N(x,y) so we conclude that this is first order differential equation.


 * {| style="width:100%" border="0"

$$F\left ( x,y \right )=\left ( 4*x^7+sin(y) \right )+\left ( x^2*y^3 \right )\frac{\mathrm{d} y}{\mathrm{d} x}=0\displaystyle $$ (2.2)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }


 * {| style="width:100%" border="0"

$$F\left ( x,(\alpha u+\beta v) \right )=\left ( 4*x^7+sin(\alpha u+\beta v) \right )+\left ( x^2*(\alpha u+\beta v)^3 \right )(\alpha \frac{\mathrm{d} u}{\mathrm{d} x}+\beta \frac{\mathrm{d} v}{\mathrm{d} x})=0\displaystyle $$ (2.3)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }


 * {| style="width:100%" border="0"

$$F\left ( x,\alpha u \right )+F\left ( x,\beta v \right )=\left ( 4*x^7+sin(\alpha u) \right )+\left ( x^2*(\alpha u)^3 \right )\alpha \frac{\mathrm{d} u}{\mathrm{d} x}+\left ( 4*x^7+sin(\beta v) \right )+\left ( x^2*(\beta v)^3 \right )\beta \frac{\mathrm{d} v}{\mathrm{d} x} \displaystyle $$ (2.4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$F\left ( x,\alpha u \right )+F\left ( x,\beta v \right )\neq F\left ( x,(\alpha u+\beta v) \right )\displaystyle$$ (2.5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

=Problem3:Plot Homogeneous Solutions And Show independence=

Given

 * {| style="width:100%" border="0"

$$y_{H}^{1}(x)=x\displaystyle $$ (3.1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$y_{H}^{2}(x)=\frac{x}{2}*log\left ( \frac{(1+x)}{(1-x)} \right )-1 \displaystyle $$ (3.2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Find
Plot $$ y_{H}^{1}(x)$$and$$y_{H}^{2}(x)$$ for visual introduction.

Show that we can not derive $$ y_{H}^{1}(x)$$ from multiplying $$y_{H}^{2}(x)$$ with $$\alpha$$

Solution
For plotting we define the interval of x between -1 to 1 to avoid imaginary parts of $$y_{H}^{2}(x)$$

 Matlab Code: 

Since for any X

$$ y_{H}^{1}(x)-\alpha*y_{H}^{2}(x)=x-\alpha *\frac{x}{2}*log\left ( \frac{(1+x)}{(1-x)} \right )-1\neq 0 $$ we can sat that $$ y_{H}^{1}(x) $$ and $$ y_{H}^{2}(x)$$ are independent homogeneous solutions.

=Problem 4=

Given

 * {| style="width:100%" border="0"

$$ \phi (x,y)=x^{2}y^{3/2}+log(x^{3}y^{2}) \displaystyle $$ (4.1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Find
$$ F(x,y,{y}')=\frac{\partial \phi (x,y)}{\partial x} \displaystyle $$, verify that $$ F(x,y,{y}') \displaystyle $$ is an exact non-linear ODE and invent three more non-linear ODE's

Solution
$$ F(x,y,{y}')=\frac{\partial \phi (x,y)}{\partial x}=M(x,y)+N(x,y){y}' \displaystyle $$

where:
 * {| style="width:100%" border="0"

$$ M=\frac{\partial \phi }{\partial y} \displaystyle $$ (4.2) and
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ N=\frac{\partial \phi }{\partial x} \displaystyle $$ (4.3) Solving for M and N we get:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ M=2xy^{3/2}+\frac{3}{xlog10} \displaystyle $$ (4.4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ N=\frac{3}{2}x^{2}y^{1/2}+\frac{2}{ylog10} \displaystyle $$ (4.5) Plugging into the equation we get:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ F(x,y,{y}')=(2xy^{3/2}+\frac{3}{xlog10})+(\frac{3}{2}x^{2}y^{1/2}+\frac{2}{ylog10}){y}' \displaystyle $$ (4.6) In order for an equation to be an exact N1-ODE the following 2 conditions must be satisified:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Condition 1
 * {| style="width:100%" border="0"

$$ M(x,y)+N(x,y){y}'=0 \displaystyle $$ (4.7) Condition 2
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ M_{y}(x,y)=0=N_{x}(x,y) \displaystyle $$ (4.8)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

For condition 1:
 * {| style="width:100%" border="0"

$$ M(x,y)+N(x,y){y}'=0 \displaystyle $$ (4.9) Using $$ M(x,y) \displaystyle $$ and $$ N(x,y) \displaystyle $$ from above we can plug in and we find:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ (2xy^{3/2}+\frac{3}{xlog10})+(\frac{3}{2}x^{2}y^{1/2}+\frac{2}{ylog10}){y}'=0 \displaystyle $$ (4.10) For Condition 2:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \frac{\partial M(x,y)}{\partial y}=\frac{\partial N(x,y)}{\partial x} \displaystyle $$ (4.11)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \frac{\partial M(x,y)}{\partial y}=3xy^{1/2} \displaystyle $$ (4.12)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \frac{\partial N(x,y)}{\partial x}=3xy^{1/2} \displaystyle $$ (4.13) Therefore:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \frac{\partial M(x,y)}{\partial y}=\frac{\partial N(x,y)}{\partial x}=3xy^{1/2} \displaystyle $$ (4.14) and $$ F(x,y,{y}') \displaystyle $$ is an exact N1-ODE.
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Three exact N1-ODE's

 Example 1 

Let's consider following equation:


 * {| style="width:100%" border="0"

$$\phi \left ( x,y \right )=\frac{x^{2}}{y^{3}}-\frac{1}{y}+c \displaystyle $$ (4.15)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$\frac{\partial \phi \left ( x,y \right )}{\partial x}=M(x,y)=\frac{2x}{y^{3}}\displaystyle $$ (4.16)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$\frac{\partial \phi \left ( x,y \right )}{\partial y}=N(x,y)=\frac{y^{2}-3x^{2}}{y^{4}}\displaystyle $$ (4.17)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Providing second condition of exactness :


 * {| style="width:100%" border="0"

$$\frac{\partial N\left ( x,y \right )}{\partial x}=\frac{\partial M\left ( x,y \right )}{\partial y}=\frac{-6x}{y^{4}}\displaystyle $$ (4.18)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Now we can write :


 * {| style="width:100%" border="0"

$$M(x,y)dx+N(x,y)dy=\frac{2x}{y^{3}}dx+\frac{y^{2}-3x^{2}}{y^{4}}dy=0\displaystyle $$ (4.19)  Example 2 
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Let's consider following equation:


 * {| style="width:100%" border="0"

$$\phi \left ( x,y \right )=e^xsiny \displaystyle $$ (4.20)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$\frac{\partial \phi \left ( x,y \right )}{\partial x}=M(x,y)=e^xsiny\displaystyle $$ (4.21)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$\frac{\partial \phi \left ( x,y \right )}{\partial y}=N(x,y)=-e^xcosy\displaystyle $$ (4.22)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Providing second condition of exactness :


 * {| style="width:100%" border="0"

$$\frac{\partial N\left ( x,y \right )}{\partial x}=\frac{\partial M\left ( x,y \right )}{\partial y}=-e^xcosy\displaystyle $$ (4.23)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Now we can write :


 * {| style="width:100%" border="0"

$$M(x,y)dx+N(x,y)dy=e^xsinydx+-e^xcosydy=0\displaystyle $$ (4.24)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

 Example 3 

Let's consider following equation:


 * {| style="width:100%" border="0"

$$\phi \left ( x,y \right )=x^{1/2}lny \displaystyle $$ (4.25)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$\frac{\partial \phi \left ( x,y \right )}{\partial x}=M(x,y)=\frac{x^{-1/2}lny}{2}\displaystyle $$ (4.26)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$\frac{\partial \phi \left ( x,y \right )}{\partial y}=N(x,y)=\frac{x^{1/2}}{y}\displaystyle $$ (4.27)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Providing second condition of exactness :


 * {| style="width:100%" border="0"

$$\frac{\partial N\left ( x,y \right )}{\partial x}=\frac{\partial M\left ( x,y \right )}{\partial y}=\frac{x^{-1/2}}{2y}\displaystyle $$ (4.28)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Now we can write :


 * {| style="width:100%" border="0"

$$M(x,y)dx+N(x,y)dy=\frac{x^{-1/2lny}}{2}dx+\frac{x^{1/2}}{y}dy=0\displaystyle $$ (4.29)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

=References=