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Homework two: Problem 3
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! '''

Homework two: Problem 7
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:

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

Homework two: Problem in 8
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:

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

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

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

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

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

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

Homework Three: Problem 11
Problem Statement and Given: 

From (| p.17-4) obtain equation 2 from p.17-3

$$Z(x)=\frac{c}{u_{1}^2}\exp\left(-\int^x a_1(s)ds\right)$$

using the integrator factor method.

Problem solution :

Starting from equation 1 from p.17-3.

$$u_{1}(x) Z'+\left[a_{1}u_{1}(x)+2u_{1}' (x) \right]Z=0 $$

$$\mu(x)\left[Z'+\left[a_{1}+\frac{2u_{1}' (x)}{u_{1}(x)} \right]Z=0 \right]$$

$$\mu(x)Z'+\mu(x)\left[a_{1}+\frac{2u_{1}' (x)}{u_{1}(x)} \right]Z=0 $$

$$\frac{d}{dx}\left[\mu(x)Z(x) \right]=\mu(x)Z'(x)+\mu'(x)Z$$

Let $$\mu'(x)= \mu(x)\left[a_{1}+\frac{2u_{1}' (x)}{u_{1}(x)} \right]$$

$$\frac{\mu'(x)}{\mu(x)}= \left[a_{1}+\frac{2u_{1}' (x)}{u_{1}(x)} \right]$$

Integrating this will give:

$$\mu(x)=B exp \left[\int_{}^{x}a_{1}(s)ds+ln\left|u_{1} \right|^{2} \right]$$

$$\mu(x)Z'+\mu'(x)Z=0$$

$$\int_{}^{}\mu(x)*Z(x)=\int_{}^{}0$$

$$\mu(x)Z(x)=K$$

$$ B exp \left[\int_{}^{x}a_{1}(s)ds+ln\left|u_{1} \right|^{2} \right]Z(x)=K $$

$$Z(x)= C exp \left[-\int_{}^{x}a_{1}(s)ds-ln\left|u_{1} \right|^{2} \right]$$

$$Z(x)= \frac{C}{u_{1}^{2}} exp \left[-\int_{}^{x}a_{1}(s)ds \right]$$

Homework Three: Problem 12
Problem Statement and Given: 

From (| p.18-1), develop reduction of order method using the following algebraic options

$$y(x)=U(x)\pm u_1 (x)$$

$$y(x)=\frac{U(x)}{u_1 (x)}$$

$$y(x)=\frac{u_1 (x)}{U(x)}$$

Problem solution :

Part a :

$$y(x)=U(x)\pm u_1 (x)$$

$$y' (x)=U' (x)\pm u_1 ' (x)$$

$$y (x)=U (x)\pm u_1 '' (x)$$

$$ (U (x)\pm u_1  (x))+a_1(U' (x)\pm u_1 ' (x))+a_0(U(x)\pm u_1 (x))=0$$

$$ (U (x)+a_1 U'(x) + a_0 U(x))\pm ((u_1(x)+a_1 u_1'(x)+a_0 u_1(x))=0$$

The reduction of order method can not be implemented since U is not missing.

Part b :

$$y(x)=\frac{U(x)}{u_1 (x)}$$

$$y' (x)= -\frac{U(x)}{u_{1}^{2}(x)}u_{1}'(x)+\frac{U'(x)}{u_{1}(x)}$$

$$y (x)= U(x)\left[-\frac{u_{1}(x)}{u_{1}(x)^{2}}+2\frac{u_{1}'(x)^{2}}{u_{1}(x)^{3}} \right]-2\frac{U'(x)u_{1}'(x)}{u_{1}(x)^{2}}+\frac{U''(x)}{u_{1}(x)}$$

$$\left[U(x)\left[-\frac{u_{1}(x)}{u_{1}(x)^{2}}+2\frac{u_{1}'(x)^{2}}{u_{1}(x)^{3}} \right]-2\frac{U'(x)u_{1}'(x)}{u_{1}(x)^{2}}+\frac{U(x)}{u_{1}(x)} \right]+a_{1}\left[-\frac{U(x)}{u_{1}^{2}(x)}u_{1}'(x)+\frac{U'(x)}{u_{1}(x)}\right]+a_{0}\left[\frac{U(x)}{u_{1}(x)} \right]=0$$

The reduction of order method can not be implemented since U is not missing.

Part c :

$$y(x)= \frac{ u_1 (x)}{ U (x) } $$

$$y' (x)= -\frac{u_1(x)}{U(x)^{2}}U'(x)+\frac{u_1'(x)}{U(x)}$$

$$y (x)= u_1(x)\left[-\frac{U(x)}{U(x)^{2}}+2\frac{U'(x)^{2}}{U(x)^{3}} \right]-2\frac{u_1'(x)U'(x)}{U(x)^{2}}+\frac{u_1''(x)}{U(x)}$$

$$\left[u_1(x)\left[-\frac{U(x)}{U(x)^{2}}+2\frac{U'(x)^{2}}{U(x)^{3}} \right]-2\frac{u_1'(x)U'(x)}{U(x)^{2}}+\frac{u_1(x)}{U(x)} \right]+a_{1}\left[-\frac{u_1(x)}{U(x)^{2}}U'(x)+\frac{u_1'(x)}{U(x)}\right]+a_{0}\left[\frac{ u_1 (x)}{ U (x) } \right]=0$$

The reduction of order method can not be implemented since U is not missing.

Homework Three: Problem 13
Problem Statement and Given: 

From (| p.18-1), Find $$u_{1}(x)$$ and $$ u_{2}(x)$$ of equation 1 on p.18-1 using 2 trial solutions:

$$ y=ax^b$$

$$ y=e^{rx}$$

Compare the two solutions using boundary conditions $$y(0)=1$$ and $$ y(1)=2$$ and compare to the solution by reduction of order method 2. Plot the solutions in Matlab.

Problem solution :

Part 1 :

Starting with the equation:

$$(1-x^{2})y''-2xy'+2y=0 $$

Using the trial solution given:

$$ y=ax^{b}$$

$$ y'=abx^{b-1}$$

$$ y''=a(b-1)x^{b-2}$$

The equation becomes:

$$ (1-x^{2})b(b-1)ax^{b-2}-2xbax^{b-1}+2ax^{b}=0$$

Now divide by: $$ ax^{b}$$

$$ (b^{2}-b)(x^{-2}-1)-2(b-1)=0$$

It is obvious that at x=0 the equation would be undefined unless b^2-b=0 which would make b=1 and at x=1 it is obvious too that b=1.

$$ y(x)=ax$$

The Boundary Conditions for this problem are y(0)=1 and y(1)=2 which only prove that $$ y=ax^{b}$$ cannot be a solution to this problem.

Part 2 :

Starting with the equation:

$$(1-x^{2})y''-2xy'+2y=0 $$

Using the trial solution given:

$$ y= exp(rx)$$

$$ y'=r exp(rx)$$

$$ y''=r^{2}exp(rx) $$

The equation becomes:

$$ (1-x^{2}) r^{2}exp(rx)-2x r exp(rx)+2 exp(rx)=0$$

Now divide by: $$ exp(rx)$$

$$ (1-x^{2}) r^{2} -2(x r -2) =0$$

It is obvious that at x=0 the value of $$r=\pm \sqrt[]{-2} $$ and at x=1 the value of r= 1.

The Boundary Conditions for this problem are y(0)=1 and y(1)=2 which only prove that $$ y= exp(rx)$$ cannot be a solution to this problem since the values for r are different for every x value.

HW4: Problem 2
Problem Statement and Given: 

Solve problem 1.1b page 28 from King book

Show that the function $$ u_{1}$$ is a solution of the differential equation given below. Use the reduction of order method to find the second independent solution, $$ u_{2}$$.

$$ xy''+2y'+xy=0$$

$$ u_{1}=x^{-1}sin(x)$$

Solution:

Find the first and second derivative of $$ u_{1}$$ and plug them into the equation to check if the equation is equal to zero.

$$ u_{1}'=x^{-1}cos(x)-x^{-2}sin(x)$$

$$ u_{1}''=-x^{-1}sin(x)-x^{-2}cos(x)-x^{-2}cos(x)+2x^{-3}sin(x)$$

$$ xu_{1}''+2u_{1}'+xu_{1}=0$$

$$ 2x^{-2}sin(x)-sin(x)-2x^{-1}cos(x)+2x^{-1}-2x^{-2}sin(x)+sin(x)=0$$

$$0=0$$

The function $$ u_{1}$$ is a solution of the differential equation given.

Now, we use the reduction of order method to find the second independent solution, $$ u_{2}$$.

Put the equation in the form:

$$ y''+a_{1}(x)y'+a_{0}y=0$$

$$ y''+\frac{2}{x}y'+y=0$$

From the notes in class, slides 17-1 to 17-4, we get the formula:

$$ u_{2}=u_{1}\int_{}^{x} \frac{1}{(u_{1}(t))^{2}}exp(-a_{1}(t))dt$$ Where: $$ -a_{1}(t)=-\int_{}^{t} a(s)ds $$

$$ -a_{1}(t)=-\int_{}^{t} \frac{2}{s}ds = -2 ln (t)=ln(t)^{-2}$$

$$ u_{2}=x^{-1}sin(x)\int_{}^{x} \frac{1}{(t^{-1}sin(t))^{2}}exp(ln t^{-2})dt$$

$$ u_{2}=x^{-1}sin(x)\int_{}^{x} \frac{1}{t^{-2}sin^{2}(t)}t^{-2}dt$$

$$ u_{2}=x^{-1}sin(x)\int_{}^{x} \frac{1}{ sin^{2}(t)} dt$$

$$ u_{2}=x^{-1}sin(x)(-cot(x)) $$

$$ u_{2}=-x^{-1}cos(x) $$

HW4: Problem 4
Problem Statement: 

Describe in words, step by step, the method of attack given only the homogeneous or non-homogeneous L2. ODE. VC.

Solution:

Answer: Try reduction of method order 0, then order 1, and later integral factor method. If they do not work, use trial solution method. If both of the roots cannot be determine we can use the undetermined factor method to find u2 knowing u1. For the non-homogenous case solve the homogeneous solution and then solve the particular solution using variations of parameters.

HW6: Problem 2
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HW6: Problem 3
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Homework Six: Problem 12
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