User:Egm6321.f09.team7/HW1

Homework one: Problem one
Given : fucnctionfunction F is a a function of Y and t where Y is also a function of t  Find : Using normal methods find the first and second total time derivatives of F     Function F $$F = f(Y(t),t)$$ Problem Solution: The first step is to find the first total time derivative for F using the chain rule (Y is implicitly a function of t and will no longer be noted as such): $$\frac{dF}{ds} = (Y,t) * \dot{Y }(t)$$ and, $$\frac{dF}{dt} = (Y,t)$$

This gives a final form for the first time derivative of, $$F,s (Y(t)*\dot{Y } + F,t (Y(t))$$

 Missing a parenthesis here, and subscripts are obtained in math mode using _{}.--Egm6321.f09.TA 03:51, 24 September 2009 (UTC)

The next step is to once again use the chain rule and find the second total time derivitive using the notation first developed in class to specify each of the parts,

$$F,s (Y,t) * \frac d{dt} \dot{Y } = F,s (Y,t)\ddot{Y } $$ $$F,ss (Y,t) * \dot{Y } = (Y,t)\dot{Y }*\dot{Y }\to(\dot{Y })^2 $$ $$F,st (Y,t) * \dot{Y } = (Y,t)\dot{Y } $$ $$F,ts (Y,t) = (Y,t)\dot{Y } $$ $$F,tt (Y,t) = (Y,t) $$

Results :adding all of the terms together gives a final equation for the second total time derivative that has the following form:

$$F,s (Y,t)\ddot{Y } + F,ss(Y,t)*(\dot{Y })^2 +F,st 2 * ((Y,t)\dot{Y }) +F,tt (Y,t)$$

Homework one: Problem two
Problem statement : Derive equation 2 given in class on page 4-2 of notes using integration factor method

Equation 2 given in class page 4-2

$$p(x)=Ae^ +x-1$$

Problem Solution:

Starting with the equations given in class reduce the second order ODE with missing dependent variable by a first order ODE with dependent variable p knowing that:

$$p(x)=y'(x)$$

the second order ODE is

$$y''+y'=x$$

and the new ODE is

$$p'+p=x$$

The method of integration factor requires a first order equation like the one above to be multiply by function u(x).

$$(p'+p)u(x)=xu(x)$$

The above equation becomes:

$$u(x)p'+u(x)p=xu(x)$$

Lets take a look at the derivative of:

$$[\frac{d}{dx}(u(x)*p)]=u(x)* [\frac{d}{dx}( p)] +p*[\frac{d}{dx}(u(x))]$$

This equation is identically to $$u(x)p'+u(x)p=xu(x)$$ as long as $$[\frac{d u(x)}{dx}]/u(x)=1$$ which means that:  Your motivation here is a little off. Multiply the original equation through by $$u(x)$$ and then enforce the exactness criteria for the new $$\overline{M},\overline{N}$$.--Egm6321.f09.TA 03:53, 24 September 2009 (UTC)

$$[\frac{d }{dx}ln|u(x)|]=1$$

After integrating both sides:

$$ln|u(x)|=x+C$$

Using the above method we find the value of u(x).

$$u(x)=Ce^x$$

Since we do not need the most general solution we can set the value of c=1.

$$u(x)=(1)*e^x$$

Now the integration factor u(x) can be replace for its real value:

$$e^x*p'+e^x*p=x*e^x$$

Integrating both sides of the following equation will give us the value of p:

$$[\frac{d}{dx}(u(x)*p)]=x*e^x$$

$$ u(x)*p =(x-1)*e^x+A$$

$$ e^x*p =(x-1)*e^x+A$$

After dividing the expression by e^x we get the value of p:

$$ p = Ae^ +x-1$$

Since we set p equal to the derivative of y, we know already that integrating p will give us the value of y:

$$ y = -Ae^ +x^2/2-x+B$$

Homework one: Problem three
Problem statement : Show that the 1st order ODE below is nonlinear

Equation 1 given in class page 4-3

$$(2*x^2+y^1⁄2)+x^5y^3y' =0$$

Problem Solution:

Starting with the definition of linearity from page 3-3 of the notes:

$$L( \alpha*u+\beta*v ) =\alpha*L(u)+\beta*L(u)$$

Using the above equation we get:

$$(2*x^2+(.)^1⁄2)+x^5(.)^3[\frac{d}{dx}(.)] =0$$

We know the equation is linear if it satisfies the test for linearity previously shown.

$$(2*x^2+(\alpha*u+\beta*v)^1⁄2)+x^5(\alpha*u+\beta*v)^3[\frac{d}{dx}(\alpha*u+\beta*v)] \neq \alpha(2*x^2+(u)^1⁄2+x^5(u)^3u')+ \beta(2*x^2+(v)^1⁄2+x^5(v)^3v') $$

The equation above failed the test because the left hand side is not equal to the right hand side. Equation (1) of page 4-3 failed the test and therefore it is nonlinear.  This is indeed correct, but your solution is difficult to read. You should take both equations and simplify them separately. Then it is clear that they are not equivalent. In their current form it is not readily obvious that they couldn't be made equivalent with some mathematical gymnastics.--Egm6321.f09.TA 03:57, 24 September 2009 (UTC)  You should define your differential operator $$L$$ as you have done, and explicitly state that you are trying to show that $$L(\alpha u+\beta v)\neq L(\alpha u)+L(\beta v)$$--Egm6321.f09.TA 03:57, 24 September 2009 (UTC)

Homework one: Problem four
Problem statement : Using theroems discussed in class show that equation given on Page 4-3 of notes is non-linear Example NonLinear 1st Order ODE $$x^2y^5+6(y')^2=0$$ Problem Solution: Using the method described in meeting 3, Define D(.) as: $$D(.)=x^2(.)^5+6[\frac{d}{dx}(.)]^2$$ then $$D(y)=x^2(y)^5+6[\frac{d}{dx}(y)]^2$$

It is linear If, $$D(\alpha*y_1+\beta*y_2) = \alpha*D(y_1) + \beta*D(y_2)$$ plugging in we see that,

$$x^2(\alpha*y_1+\beta*y_2)^5+6[\frac{d}{dx}(\alpha*y_1+\beta*y_2)]^2 \neq \alpha*(x^2y_1^5+6(y_1')^2) + \beta*(x^2y_2^5+6(y_2')^2)$$

Results : equation is shown to be nonLinear.  Again, simplify the two expressions so that it is clear that they are not equivalent. This problem however is clearer in terms of explanation than problem 3 (see note above).--Egm6321.f09.TA 03:59, 24 September 2009 (UTC)

Homework one: Problem five
Problem statement :(6-1): Generating exact nonlinear 1st order ODEs: Let $$ \Phi(x,y)=6x^{4}+2y^{3/2}$$, then $$ M=\Phi_{x}$$ $$ N=\Phi_{y}$$ Complete the details and invent 3 more examples.

Problem solution Just invent $$\phi$$: Completing details:

$$\phi(x,y)=6x^4+2y^{\frac{3}{2}}$$ $$M=\phi_x$$ $$N=\phi_y$$ $$M+Ny'$$ $$24x^3+3\sqrt{y}y'=0$$ 1st invention $$\phi(x,y)=e^x+e^y$$ $$e^x+e^yy'=0$$ 2nd invention $$\phi(x,y)=\sin(x)+\sin(y)$$ $$\cos(x)+\sin(y)y'=0$$ 3rd invention $$\phi(x,y)=\ln(x)+\ln(y)$$ $$\frac{1}{x}+\frac{y'}{y}=0$$

 Nice work.--Egm6321.f09.TA 04:03, 24 September 2009 (UTC)