User:Egm6321.f09.team7/HW2

Homework Two: Problem One
Problem statement :(7-1): Complete Details to obtain h(y): Given : $$h_xN=0$$ $$$$ $$\frac{}{}-h_yM+h(N_x-M_y)=0$$ $$\frac{h_y}{h}=\frac{1}{M}(N_x-M_y)=-g(x)$$

Problem solution : $$\int\frac{h_y}{h}dy=-\int g(y) dy$$ $$ln(h)=-\int g(y)dy$$ $$h(y)=exp[-\int g(y) dy]$$ Where: $$-g(y)=\frac{1}{M}(N_x-M_y)$$

 Your solution is correct, but it cannot stand on its own legs. Where did the first expression come from? We have chosen the case where $$ h_xN=0$$. --Egm6321.f09.TA 05:03, 28 September 2009 (UTC)

Homework Two: Problem Two
Problem statement :(8-2,8-3): Apply Non-Homog L1.ODE.VC : Given : $$y'+\frac{1}{x}y=x^2$$

Problem solution : $$y'+(\frac{1}{x}y-x^2)=0 $$ $$h(x)=e^{\int\frac{1}{x}dx} $$ $$h(x)=x $$ Multiply by $$h(x),$$ $$xy'+y=x^3 $$ $$\frac{d(xy)}{dx}=x^3 $$ $$xy=\int x^3 dx $$ $$xy=\frac{x^4}{4}+C_1 $$ $$y=\frac{x^3}{4}+\frac{C_1}{x} $$

 Clear and concise. --Egm6321.f09.TA 05:03, 28 September 2009 (UTC)

Homework two: Problem Three
Problem statement : Show that Eq. 3 is exact in page 9-2

Equation 3 given in class page 9-2

$$ 1/2*x^2*y'+(x^4*y+10)=0$$

Problem Solution:

Starting with the formula given in class named Eq. 2 page 9-1 which is used to test for the exactness of None linear first order ODEs with varying coefficients we can get:

$$b(x)y' +(a(x)y+k)=0$$

$$ y' +a(x)y/b(x)+k/b(x)=0$$

Likewise we can use the same analogy with equation 3:

$$y'+(2(x^4*y+10))/x^2=0$$

Now we can let:

$$N(x,y)=1$$

$$M(x,y)=(2(x^4*y+10))/x^2$$

Then we can use:

$$1/N(x,y)*(N_x(x,y)-M_y(x,y))=-f(x)$$

Since N(x,y)=1 then $$N_x=0$$ and $$f(x)=M_y(x,y)$$. This means that, h(x)=exp(ʃf(x)dx) h(x)=exp(ʃ$$M_y$$dx)

$$h(x)=exp(2/3*x^3)$$

Now we can multiply Eq. 3 by h(x),

$$ exp(2/3*x^3)y'+2x^2* exp(2/3*x^3)* y+ 20*exp(2/3*x^3)/x^2=0$$

Now:

$$\bar{N}(x,y)=h(x)*N(x,y)= exp(2/3*x^3)$$

$$\bar{M}(x,y)=h(x)*M(x,y)= 2x^2* exp(2/3*x^3)* y+ 20*exp(2/3*x^3)/x^2$$

In order for an ODE to be exact $$\bar{N_x}=\bar{M_y}$$

$$2x^2* exp(2/3*x^3)= 2x^2* exp(2/3*x^3)+0$$ TRUE!

 Good. If you use more words for explanation, you can condense the solution with less math. This is very clear however. --Egm6321.f09.TA 05:03, 28 September 2009 (UTC)

Homework Two: Problem Four
Problem statement :: Show that Equation 8 from page 9-3 is an exact NL1ODE: Given : Equation 8 9-3: $$\frac13x^3y^4\dot{Y}+5x^3+2(\frac15Y^5)=0$$ $$$$

Problem solution : First step is to recognize that the equation is in the form: $$b(x)C(y)\dot{Y}+a(x)c(y)=0$$ Then we can divide through by $$b(x)C(y)$$: $$\dot{Y}+\frac{a(x)c(y)}{b(x)C(y)}=0$$ or $$\dot{Y}+\frac{5x^3+2}{\frac13x^3}*\frac{\frac15Y^5}{Y^4}=0\Rightarrow \dot{Y}+\frac{5x^3+2}{\frac13x^3}*\frac{Y}{5}=0$$ This gives a form similar to page 6-3 eqn3 : $$h(x)=exp[\frac15\int\frac{5x^3+2}{\frac13x^3}]dx$$ integrating and reducing leads us to: $$h(x)=exp[3x-\frac3{5x^2}]$$ recalling that $$M(x,y)=[3+\frac6{5x^3}]Y$$ and that $$N(x,y)=1$$ after we divided through in the first step we can multiply through by our $$h(x)$$to find that our new: $$\bar{M}(x,y)=M(x,y)*h(x)$$and $$\bar{N}(x,y)=1*h(x)$$ so to prove the exactness we need to meet the condition: $$\bar{N}x=\bar{M}y$$ Finding the partial derivatives gives us: $$\bar{N}x=exp[3x-\frac3{5x^2}]*[3+\frac6{5x^3}]$$ and $$\bar{M}y=exp[3x-\frac3{5x^2}]*[3+\frac6{5x^3}]$$  Thus proving that $$\bar{N}x=\bar{M}y$$ and that this is an exact NL1ODE 

 Good write up.

Homework Two: Problem Five
Problem statement :: Show that the second exactness condition for xyy&apos;&apos; + x(y')2 + yy' = 0 is satisfied: Given : Equation 8 9-3: $$xyy'' + x(y')2 + yy' = 0$$ and Second exactness condition: $$Fxp+P*Fyp+2Fy=Gpp$$ $$$$

Problem solution : First step is to identify both F and G from the original function (F being the equation accompanying the second derivative of Y): so from the original problem F is given as: $$XY$$ and from the original problem G is given as: $$x(\dot{Y})^2+Y\dot{Y}$$ The next step will be to identify the necessary partial derivatives to fulfill the second exactness condition: $$Fx=Y$$ $$Fxp=0$$ $$Fy=x$$ $$Fyp=0$$ $$Gp=2x\dot{Y}+Y$$ $$Gpp=2x$$ Finally we the above partials should be inserted into the second exactness condition: $$0+0+0+2X=2X\rightarrow2X=2X$$  This proves that the second condition is met 

 This is the second equation in the second exactness criteria. Where is the proof of the first equation? --Egm6321.f09.TA 05:03, 28 September 2009 (UTC)

Homework two: Problem Six
Problem statement : Derive Eq.(5) p.10-2 by differentiating Eq.(3) p.10-1 with respect to p=y'

Equation (5) given in class page 10-2

$$f_{xp}+pf_{yp}+2f_{y}=g_{pp}$$

Equation (3) given in class page 10-2

$$g(x,y,p)=\phi_{x}+\phi_{y}p$$

Problem Solution:

$$g=\phi_{x}+\phi_{y}p$$

$$g_{pp}=\phi_{xpp}+(\phi_{yp}p)_{p}+\phi_{yp}$$

$$g_{pp}=\phi_{xpp}+\phi_{ypp}p+2\phi_{yp}$$

Note: We know that:

$$f(x,y,p)=\phi_{p}(x,y,p)$$

$$\phi_{pp}=f_{p}$$

$$\phi_{ppy}=f_{py}$$

$$\phi_{ppx}=f_{px}$$

Since $$\phi_{xy}=\phi_{yx}$$

$$\phi_{xpp}=f_{xp}$$

The final solution gives us:

$$g_{pp}=f_{xp}+f_{py}p+2f_{y}$$

 Clear and concise. --Egm6321.f09.TA 05:03, 28 September 2009 (UTC)

Homework two: Problem Seven
Problem statement : Use $$\phi_{xy}=\phi_{yx}$$ to obtained Eq.(4) p.10-2. [Use Eq.(3) & Eq.(4) from p.11-4 & p.12-1] Equation (3) given in class page 11-4

$$\phi_{x}=g-p(g_{p}-f_{x})+p^{2}f_{y} $$

Equation (4) given in class page 12-1

$$ \phi_{y}=g_{p}-f_{yp}-f_{y} $$

Equation (4) given in class page 10-2

$$ f_{xx}+2pf_{xy}+p^{2}f_{yy}=g_{xp}+pg_{yp}-g_{y} $$

Problem Solution:

$$\phi_{y}=g_{p}-f_{y}p-f_{x}$$

$$\phi_{x}=g-p(g_{p}-f_{x})+p^{2}f_{y}$$

$$\phi_{yx}=g_{px}-f_{yx}p-f_{xx}$$

$$\phi_{xy}=g_{y}-p(g_{py}-f_{xy})+p^{2}f_{yy}$$

$$\phi_{yx}=\phi_{xy}$$

$$g_{xp}+pg_{yp}-g_{y}=f_{xx}+2pf_{xy}+p^{2}f_{yy}$$

 You need some explanation between your math. This is 6 straight lines of math without words to connect them together. --Egm6321.f09.TA 05:03, 28 September 2009 (UTC)

Homework Two: Problem Eight
Problem statement :: Verify exactness condition 2, equations 4&5 on p.(10-2). for the differential equation 8x5y'y&apos;&apos; + 2x2y' + 20x4(y')2 + 4xy = 0: Given : Equation to be solved: $$8x5y'y'' + 2x2y' + 20x4(y')2 + 4xy = 0$$ and Second exactness condition:

$$Fxx+2P*Fxy+P^2*Fyy=Gxp+P*Gyp-Gy$$

$$Fxp+P*Fyp+2Fy=Gpp$$

$$$$

Problem solution : First step is to identify both F and G from the original function (F being the equation accompanying the second derivative of Y): so from the original problem F is given as: $$8x^5\dot{Y}$$ and from the original problem G is given as: $$2x^2P+20x^4P^2+4xY$$ The next step will be to identify the necessary partial derivitives to fulfill the second exactness condition: $$Fx=40x^4P$$ $$Fxp=40x^4$$ $$Fxx=160x^3P$$ $$Fxy=0$$ $$Fy=0$$ $$Fyy=0$$ $$Fyp=0$$ $$Gp=2x^2+40x^4P$$ $$Gx=4xP+80x^3P^2+4Y$$ $$Gy=4x$$ $$Gpp=40x^2$$ $$Gxp=4x+160x^3P$$ $$Gyp=0$$

Finally we the above partials should be inserted into the second exactness condition:

condition 2 part 1: $$160x^3P+0+0=4x+160x^3P+0-4x\rightarrow160x^3P= 160x^3P$$ condition 2 part 2: $$40x^4+0+0=40x^4\rightarrow40x^4=40x^4$$

 This proves that the second condition is met 

 good. --Egm6321.f09.TA 05:03, 28 September 2009 (UTC)

Homework Two: Problem Nine
Problem statement :(12-3): Verify the exactness of the ODE

Given : Equation 1 12-3: $$\sqrt{x}\frac{d^2y}{dx^2} + 2x\frac{dy}{dx} + 3y = 0$$

Problem solution : Condition 1 is met! $$f(x,y,p) = \sqrt{x}$$ $$g(x,y,p) = 2xp + 3y$$ Condition 2 Test: $$g_{pp} = f_{xp} + pf_{yp} + 2f_{y}$$ $$0 = 0 + p\cdot0 + 0$$ TRUE! $$f_{xx} +2pf_{xy} + p^2f_{yy} = g_{xp} + pg_{yp} - g_{y}$$ $$\frac{-1}{2}\frac{1}{2}x^\frac{-3}{2} + 0 + 0 = 2 + 0 - 3$$ $$\frac{-1}{4}x^\frac{-3}{2} = -1$$ FALSE!

It is not exact currently since condition 2 is not met; however, it might be exactly integrable!  good work. --Egm6321.f09.TA 05:03, 28 September 2009 (UTC)

Contributing Authors
Egm6321.f09.team7.benedict 03:07, 22 September 2009 (UTC) Egm6321.f09.team7.mm 16:00, 22 September 2009 (UTC) Egm6321.f09.team7.hua 02:11, 23 September 2009 (UTC) User:Emg6321.f09.blanco 11:17, 23 September 2009 (UTC)