User:Egm6321.f12.team1.cla/report3

=Problem 3.1* Integration constant necessity=

Link to lecture notes

Find:
Show that integration constant $$k_{1}$$ is insignificant

Solution:
By plugging in equation ($$) into Equation ($$) we can show that $$k_{1}$$ cancels out


 * $$y(x)=\frac{1}{ \exp \left (\int_{x} a_{0}(x)dx+k_{1}\right )} \left ( \int_{x} exp\left ( \int_{x} a_{0}(x)dx+k_{1}\right )b(x)dx+k_{2}\right )$$

We can extract the constants from the exponential within the integral as follows


 * $$y(x)=\frac{1}{exp(k_{1})exp\left ( \int_{x} a_{0}(x)dx\right )}\left ( exp(k_{1})\int_{x} exp\left ( \int_{x} a_{0}(x)dx\right )b(x)dx+k_{2}\right )$$

Then they can be cancelled out


 * $$y(x)=\frac{1}{\cancel{exp(k_1)}exp\left ( \int_{x} a_{0}(x)dx\right )}\left ( \cancel{\exp(k_1)}\int_{x} \exp\left ( \int_{x} a_{0}(x)dx\right )b(x)dx+k_{2}\right )$$

=Problem 3.2* Show similarity between the solutions from King and from lecture=

Link to lecture notes

Given:
 King's solution format for differential equations

Equation for the solution of differential equations from lecture

With h(x),

Find:
Show the similarity between King's solution to differential equations and the solution from lecture by identifying what terms in equation 3.2 and 3.3 are related to A, $$y_{H}(x)$$, and $$ y_{p}(x)$$ terms in equation 2.1.

Solution:
First let's find A and $$y_{H}(x)$$ using equation 2.3, 
 * $$ h(x)=\exp[\int^x a_{0}(s)ds+k_{1}] $$

 Simplifying h(x) using exponential properties we see, 
 * $$ h(x)=k_{1}\exp[\int^x a_{0}(s)ds] $$

 According to King pg. 512 the homogeneous solution of a differential equation takes the form, 
 * $$ y_{H}(x)=A\exp[-\int^x P(t)dt] $$

 Therefore comparing h(x) from lecture and $$ y_{H}(x)$$ from King we can see, 
 * $$ A=k_{1} and y_{H}(x)=h(x)=A\exp[-\int^x P(t)dt]$$

 with h(x) and y_{H}(x) both being the homogeneous solution of a differential equation.  Now we can prove that y(x) from lecture is the same the $$ y_{p}(x) $$.  First let's start with equation 2.2, 
 * $$ y(x)=\frac{1}{h(x)}[\int^x h(s)b(s)ds+k_{2}] $$

 where s is a dummy variable or in other words can be anything.  Plugging in King's version of h(x) yields, 
 * $$ y(x)=\frac{1}{A\exp[-\int^x P(t)dt}][\int^x exp(h(s)b(s)ds+k_{2})] $$

 Now we need to use an exponential property pointed out below, <br\>
 * $$ \exp(ab)=\exp(a)\exp(b) $$

<br\> Applying this property to equation 2.2 we get <br\>
 * $$ y(x)=\frac{1}{A\exp[-\int^x P(t)dt}][\int^x \exp(h(s)ds+k_{3})\int^x exp(b(s)ds+k_{4})] $$

<br\> where k is an arbitrary constant. <br\> Now reorganizing the terms we have, <br\>
 * $$ y(x)=A\exp[-\int^x P(t)dt][\int^x \exp(P(t)ds+k_{3})\int^x exp(b(t)ds+k_{4})ds] $$

<br\> From King we have the solution for the particular solution of a differential equation to be, <br\>
 * $$ y(x)=\exp[-\int^x P(t)dt][\int^x \exp(P(t)ds)\int^x exp(Q(s)ds)ds] $$

<br\> Seeing the similarity between the King's version and the version from lecture we can see that y(x) from lecture is the particular solution of a differential equation from King pg. 512.

=Problem 3.3* Identifying $$y_H(x)$$ from the homogeneous counterpart of the equation= Instead of identifying $$y_H(x)$$ form equation (3) p11-4 and equation (1) p11-5 solve the homogeneous counterpart of equation (3) p11-3, i.e.,


 * $$ y' + a_0(x)y = 0 $$

Find:
Identifying $$y_H(x)$$ from the homogeneous counterpart of the equation


 * $$ y' + \frac{Q(x)}{P(x)} y = \frac{R(x)}{P(x)} $$

Solution:
The homogeneous part of the equation is

Where
 * $$ y' = \frac{dy}{dx} $$

Rearranging ($$)
 * $$ \frac{Q(x)}{P(x)} y = -\frac{dy}{dx} $$

Using the variable seprable method for solving the equation.
 * $$ \int-\frac{1}{y}dy = \int\frac{Q(x)}{P(x)}dx $$

$$\Rightarrow -\log |y| = \int\frac{Q(x)}{P(x)}dx $$

$$\Rightarrow y_H = e^{\int\frac{Q(x)}{P(x)}dx}\qquad\blacksquare $$ --Pavel.bhowmik (talk) 14:12, 1 October 2012 (UTC)

=Problem 3.4* Find the integrating factor =

Find:
If ($$) is not exact, find the integrating factor $$h$$ to make it exact.

Solution:
The first exactness condition says the equation needs to be able to be in the form of equation 4.2. Since equation 4.1 was given in this form it passes the first exactness condition.

Equation ($$) shows the second exactness condition

$$\left\{\begin{matrix}M(x,y):=x^4y+10 \\N(x,y):=\frac{1}{2}x^2\,\,\end{matrix}\right.$$


 * $$M_y(x,y) = x^4$$


 * $$N_x(x,y) = x$$

We find that $$M_y(x,y) \neq N_x(x,y)$$ Therefore, ($$)is not exact.

To find the integrating factor
 * $$ \frac{h_y}{h} = -\frac{1}{N}(N_x - M_y)=-\frac{2}{x^2}(x-x^4):=n(x)$$

$$\Rightarrow h(x)=\exp\left ( \int\frac{2}{x}(1-x^3)dx \right )$$

$$\Rightarrow h(x)=\exp\left ( \frac{2}{3}x^3-2lnx \right )$$

$$\Rightarrow h(x)=\frac{1}{x^2}\exp\left ( \frac{2}{3}x^3\right )$$

=Problem 3.5* Show that an integrating factor h(x) can be found=

Given
<br\>

and

Prove that
Equation ($$) Satisfies the condition
 * $$\frac {h_x} {h} = - \frac {1} {N} (N_x - M_y) =: n(x)$$,

<br\> That is $$\frac {1} {N} (N_x - M_y) $$ is only a function of $$x$$ and an integrating factor $$h(x)$$ can be found to render it exact only if $$k_1(y)=d_1$$ (Constant). <br\> Further show that ($$) includes the following equation

<br\> as a particular case.

Solution
Given<br\> $$ \begin{array}{lcl} & N(x,y)  & = \bar {b} (x,y)c(y) \\ &         & = \left ( \int^x b(s)\mathrm{d}s+k_1(y) \right )c(y)  \\ &         & \\ \Rightarrow  & N_x(x,y) & = b(x)c(y) \\ &         & \end{array} $$ <br\> and <br\> $$ \begin{array}{lcl} & M(x,y)  & = a(x)\bar {c} (x,y) \\ &         & = a(x) \left ( \int^y c(s)\mathrm{d}s+k_2(x) \right )  \\ &         & \\ \Rightarrow  & M_y(x,y) & = a(x)c(y) \\ &         & \end{array} $$ <br\> $$ \Rightarrow  \frac {1} {N} (N_x - M_y)  = \displaystyle\frac {1} {\left ( \int^x b(s)\mathrm{d}s+k_1(y) \right ) \cancel{c(y)}} (b(x)\cancel{c(y)} -a(x)\cancel{c(y)}) $$

The only dependency on $$y$$ of the right hand side of ($$) is through the term $$k_1(y)$$. So the left hand side is independent of $$y$$ if and only if $$k_1(y)$$ is a constant. That is $$ \frac {1} {N} (N_x - M_y) $$ is independent of $$y$$ if and only if $$k_1(y) = d_1$$, a constant$$.\qquad\blacksquare$$ <br\><br\><br\> In the case that $$k_1(y) = d_1$$, a constant, then<br\>


 * $$\bar {c} (x,y) = \int^y c(s)\mathrm{d}s+k_2(x) = \int^y \mathrm{d}s+k_2(x) $$

Substituting ($$), ($$), ($$) into ($$)


 * $$\bar b(x) y' + a(x)[y+k_2(x)]=0$$<br\><br\>

$$\Rightarrow [a(x)y+a(x)k_2(x)]+\bar b(x) y'=0$$<br\><br\> $$\Rightarrow [a(x)y+k_3(x)]+\bar b(x) y'=0$$ <br\><br\>Which has exactly the same form as ($$)$$.\qquad\blacksquare$$ Pavel Bhowmik (talk) 14:16, 1 October 2012 (UTC)

=Problem 3.6* Made exact by Integration Factor Method =

Given:

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


 * style="width:100%" |
 * $$\left ( \frac{1}{3}x^3+d_{1} \right )(y^4)y'+(5x^3+2)\left ( \frac{1}{5}y^5+\sin x+d_{2}\right )=0$$
 * $$\left ( \frac{1}{3}x^3+d_{1} \right )(y^4)y'+(5x^3+2)\left ( \frac{1}{5}y^5+\sin x+d_{2}\right )=0$$

<p style="text-align:right"> (6.1)
 * }

Find:
Determine whether equation 6.1 is exact and if it is not show how it can be made exact using the integrating factor method

Solution:
The first exactness condition says the equation needs to be able to be in the form of equation 6.2. Since equation 6.1 was given in this form it passes the first exactness condition.

Equation 6.3 shows the second exactness condition

$$M_{y}(x,y)=y^4(5x^3+2)$$ $$N_{x}(x,y)=x^2y^4$$ Therefore $$M_{y}(x,y)\neq N_{x}(x,y)$$ Equation 6.1 can be made exact by multiplying by the integration factor shown in equation 6.4

Re-write equation 6.1 elimination any integration constants or functions

Equation 6.5 is in the following form $$\bar{b}(x,y)c(y)y'+a(x)\bar{c}(x,y)=0$$ $$\bar{b}(x)=\left(\frac{x^3}{3}+d_{1}\right)$$ $$c(y)=y^4$$ $$a(x)=5x^3+2$$ $$\bar{c}(x,y)=\frac{y^5}{5}+\sin x+d_{2}$$ $$b(x)=x^2$$ This equation is from lecture notes 13 pg 6 equation 2 and can be used to find the integrating factor to make equation 6.5 exact

$$n(x)=\frac{-x^2+5x^3+2}{\frac{x^3}{3}+d_{1}}$$

$$h(x)=exp\left[\int_{x}\frac{-s^2+5s^3+2}{\frac{s^3}{3}+d_{1}}ds+k\right]$$ $$h(x)=exp\left[\int_{x}\frac{-3s^2+15s^3+6}{s^3+d_{1}}ds+k\right]$$ Equation 6.5 is made exact by multiplying by the integration factor and the result is shown below

=Problem 3.7* Show that the function is exact or can be made exact and find the first integral=

Given:
<br\>

and

<br\>

<br\>

<br\>

Find:
Put the arbitrary equations $$a(x), b(x)$$ and $$c(y)$$ in form of $$M(x,y)$$ and $$N(x,y)$$. Then find the first integral $$\phi(x,y)=k $$ <br\>

Solution:
First we need to find,

<br\> Plugging $$b(x)$$ into equation ($$) yields,
 * $$\bar {b} (x,y) = \int^x sin(x)\mathrm{d}x+k_1(y)$$

Integrating we get,
 * $$\bar {b} (x,y) = cos(x)+k_1(y)$$

Now we need to find,

<br \> Plugging $$c(y)$$ into equation ($$) yields,
 * $$\bar {c} (x,y) = \int^x exp(2y)\mathrm{d}y+k_2(x)$$

Integrating we get,
 * $$\bar {c} (x,y) = \frac{1}{2}exp(2y)+k_2(x)$$

Now that we have $$\bar {b} (x,y)$$ and $$\bar {c} (x,y)$$ we can plug them into equations ($$),

which has the form,

Therefore in order to find $$\phi(x,y)=k $$ we just have to integrate the above equation,
 * $$\int \frac {d\phi(x,y(x))}{dx}=\int\left(\frac{\partial\phi(x,y)}{\partial x}+\frac{\partial\phi(x,y)}{\partial y}\frac{dy}{dx}\right)=k$$
 * $$\phi(x,y)=\int \frac {d\phi(x,y(x))}{dx}=\int^x \left(sin(x^3)dx \left(\frac{1}{2}exp(2y)+k_2(x)\right)\right)+\int^y((sin(x^3)+k_1(y))(exp(2y))dy)y'$$

Now solving the above integral gives the solution,
 * $$\phi(x,y)=\frac{1}{4}cos\left(\frac{x^4}{4}\right)\left(\frac{1}{2}exp(2y)\right)+\left(\frac{1}{4}exp(2y)\right)sin(x)=k$$

=Problem 3.8* Create a group of differential equations that can be made exact by integrating factor method=

Given:
where a,b,c are arbitrary functions

Find:
Counterpart of

that satisfies the condition

Solution:
For ($$), from the definition of $$\bar{b}$$ taking the derivatives with respect to x

Taking the derivative of ($$)with respect to y

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

$$\frac{h_y}{h}=\frac{1}{M}(N_x-M_y) = \frac{1}{c(y)\bar{a}(x,y)}(c(y)b(y)-c(y)a(x))$$
 * style="width:100%" |
 * style="width:100%" |
 * }
 * }

=Problem 3.9* Application (engineering) Motion of a particle in the air =

Given:


$$v:=||v||$$ particle velocity

$$k,n\in \mathbb{R}$$ constants

$$m\,\,$$ mass of particle

$$g\,\,$$ acceleration of gravity

Find:
Derive the euquations of motion

form a System of Coupled N1-ODEs=SC-N1-ODEs (need numerical methods)

Solution:
The sum of force along x direction equals zero, that is

$$\sum f_x=ma_x+kv^n\cos\alpha=m\frac{dv_x}{dt}+kv^n\cos\alpha=0$$

Here comes the Equation($$) The same as sum of force along y direction, which also equals zero, that is

$$\sum f_y=ma_x+kv^n\sin\alpha+mg=m\frac{dv_y}{dt}+kv^n\sin\alpha+mg=0$$

Here comes the Equation($$) Equation($$)is from definition of magnitude of a vector;

$$\displaystyle ||v||=\sqrt{(v_x)^2+(v_y)^2}$$

while equation($$) comes from trigonometry

Find:
Particular case $$\displaystyle k = 0$$ : Verify that $$\displaystyle y(x) $$ is parabola.

Solution:
Apply $$\displaystyle k = 0$$ into Equation ($$) gives

Integrate these two Equations, we get

Because

Apply ($$)&($$) into ($$)&($$)

Integrate ($$)&($$)provides the distance

Vanishing $$t$$ to get the relationship between $$x(t)$$&$$y(t)$$ gives

Observe ($$),$$x$$is the only variable and the highest order of $$x$$ is 2, all other elements are constants. The coefficient of $$x^2$$ is $$\frac{g}{2v_x0^2} \neq 0$$, therefore $$\displaystyle y(x) $$ is parabola.

Find:


Consider the case $$\displaystyle k \neq 0$$ and $$\displaystyle v_{x0} = 0$$

3.1 Is ($$) either exact or can be made exact using IFM? Find $$\displaystyle v_{y}(t)$$ and $$\displaystyle y(t) $$ for $$\displaystyle m $$ constant.

3.2 Find $$\displaystyle v_{y}(t)$$ and $$\displaystyle y(t)$$ for $$\displaystyle m = m(t)$$

Solution:
Rewrite Equation ($$)in the form of


 * When $$n=0$$

Equation ($$)is exact and gives


 * When $$n=1$$

Equation ($$)is satisfied with the first exactness condition, and $$N_t(t,v_y)=0$$ $$M_{v_y}(t,v_y)=\frac{k}{m}$$

If $$k=0$$,then $$ N_t(t,v_y)=M_{v_y}(t,v_y)=0$$ Equation $$is exact, and

If $$k\neq 0$$,then $$N_t(t,v_y)\neq M_{v_y}(t,v_y)$$ Equation $$is not exact but can be made exact using IFM, see

$$h(t)=exp[\int\frac{k}{m}dt]$$ $$v_y=\frac{1}{h(t)}\int -g h(t) dt$$


 * When $$n=2 $$

Equation ($$)is not exact because it cannot satisfied with the first exactness condition, and cannot be made exact using IFM

According to (Figure9.2), we get $$m(t)=\frac{m_1-m_0}{t_1}t\,\,,t\in [0,t_1]$$ $$m(t)=m_1\,\,\,\,\,\,\,\,,t\in (t_1,+\infty)$$

Part 3.1 $$m$$ is constant, which means $$ t\in (t_1,+\infty)$$ Equation($$) becomes
 * When $$n=0$$

Equation($$) becomes
 * When $$n=1$$

Therefore

The problem cannot be solved because Equation ($$) neither exact nor can made exact.
 * When $$n=2$$

Part 3.2 $$m$$ is not constant, which means $$ t\in [0,t_1]$$

Equation($$) becomes
 * When $$n=0$$

Therefore

Equation($$) becomes
 * When $$n=1$$

Therefore

The problem cannot be solved because Equation ($$) neither exact nor can made exact.
 * When $$n=2$$

=Problem 3.10* Prove the solution of the differential equation=

Show That
The solution of ($$) is<br\>
 * $$x(t) = [\exp\{a(t-t_0)\}]x(t_0) + \int_{t_0}^t [\exp\{a(t-\tau)\}]b u(\tau) \mathrm{d} \tau$$

Solution
We can re-write ($$) as

$$\Rightarrow -\frac {1} {N} (N_t-M_x) = - a$$<br\> So the Integrating Factor is given as
 * $$h(t) = \exp \left[ \int^t -a \; \mathrm{d}s + k_1 \right] = \exp \left[ -at + k_1 \right] $$

Multiplying both sides of ($$) with the Integration Factor, we get<br\><br\>
 * $$-e^{\displaystyle \left\{ -at + k_1 \right\}} a\,x(t)+e^{\displaystyle \left\{ -at + k_1 \right\}} \dot x(t) = e^{\displaystyle \left\{ -at + k_1 \right\}} b u(t)$$

Which is in exact form<br\> $$\Rightarrow \frac {\mathrm{d}} {\mathrm{d}t} \left (e^{\displaystyle \left\{ -at + k_1 \right\}} x(t) \right ) = e^{\displaystyle \left\{ -at + k_1 \right\}} b u(t)$$<br\><br\>

Integrating with respect to $$t$$<br\>

Using the boundary condition at $$t=t_0, x(t)=x(t_0)$$<br\>

Subtracting ($$) from ($$)
 * $$e^{\displaystyle \left\{ -at + k_1 \right\}} x(t) - e^{\displaystyle \left\{ -at_0 + k_1 \right\}} x(t_0)= \int^t e^{\displaystyle \left\{ -a\tau + k_1 \right\}} b u(\tau) \mathrm{d} \tau - \int^{t_0} e^{\displaystyle \left\{ -a\tau + k_1 \right\}} b u(\tau) \mathrm{d} \tau + k_2$$

$$\Rightarrow e^{\displaystyle \left\{ -at + k_1 \right\}} x(t) = e^{\displaystyle \left\{ -at_0 + k_1 \right\}} x(t_0) + \int_{t_0}^t e^{\displaystyle \left\{ -a\tau + k_1 \right\}} b u(\tau) \mathrm{d} \tau$$<br\> Dividing both sides by $$e^{\displaystyle \left\{ -at + k_1 \right\}}$$, we get
 * $$ x(t) = e^{\displaystyle \left\{ a(t-t_0) \right\}} x(t_0) + \int_{t_0}^t e^{\displaystyle \left\{ a(t-\tau)\right\}} b u(\tau) \mathrm{d} \tau \qquad\blacksquare$$

<br\><br\> The homogeneous solution to ($$) is found by solving the same equation, only setting the forcing function to $$0$$. So the homogeneous solution is solution to the following equation<br\>
 * $$ -a x(t) + \dot x (t) = 0$$

And the homogeneous solution is
 * $$x(t) = e^{\displaystyle \left\{ a(t-t_0) \right\}} x(t_0)$$

The particular solution is the second term in $$x(t)$$
 * $$ x(t) = \int^t e^{\displaystyle \left\{ a(t-\tau)\right\}} b u(\tau) \mathrm{d} \tau$$

Pavel Bhowmik (talk) 14:14, 1 October 2012 (UTC)

Show That
The solution of ($$) is<br\>
 * $$x(t) = \left [\exp\int_{t_0}^t a(\tau) \mathrm{d} \tau \right ]x(t_0) + \int_{t_0}^t \left [\exp\int_{\tau}^t a(s) \mathrm{d}s \right ]b u(\tau) \mathrm{d} \tau$$

Solution
We can re-write ($$) as

$$\Rightarrow -\frac {1} {N} (N_t-M_x) = - a(t)$$<br\> So the Integrating Factor is given as
 * $$h(t) = \exp \left[ \int^t -a(s) \; \mathrm{d}s + k_1 \right] $$

Multiplying both sides of ($$) with the Integration Factor, we get<br\><br\>
 * $$ -\exp \left[ \int^t -a(s) \; \mathrm{d}s + k_1 \right] a(t)\,x(t)+\exp \left[ \int^t -a(s) \; \mathrm{d}s + k_1 \right] \dot x(t) = \exp \left[ \int^t -a(s) \; \mathrm{d}s + k_1 \right] b(t) u(t)$$

Which is in exact form<br\> $$\Rightarrow \frac {\mathrm{d}} {\mathrm{d}t} \left (\exp \left[ \int^t -a(s) \; \mathrm{d}s + k_1 \right] x(t) \right ) = \exp \left[ \int^t -a(s) \; \mathrm{d}s + k_1 \right] b(t) u(t)$$<br\><br\>

Integrating with respect to $$t$$<br\>

Using the boundary condition at $$t=t_0, x(t)=x(t_0)$$<br\>

Subtracting ($$) from ($$)
 * $$ \exp \left[ \int^t -a(s) \; \mathrm{d}s + k_1 \right] x(t) - \exp \left[ \int^{t_0} -a(s) \; \mathrm{d}s + k_1 \right]  x(t_0)$$<br\>
 * $$ = \int^t \exp \left[ \int^\tau -a(s) \; \mathrm{d}s + k_1 \right] b(\tau) u(\tau) \mathrm{d} \tau -  \int^t \exp \left[ \int^\tau -a(s) \; \mathrm{d}s + k_1 \right]  b(\tau) u(\tau) \mathrm{d} \tau$$

$$\Rightarrow \exp \left[ \int^t -a(s) \; \mathrm{d}s + k_1 \right] x(t) =  \exp \left[ \int^{t_0} -a(s) \; \mathrm{d}s + k_1 \right]  x(t_0) + \int_{t_0}^t \exp \left[ \int^\tau -a(s) \; \mathrm{d}s + k_1 \right]  b(\tau) u(\tau) \mathrm{d} \tau$$<br\>

Dividing both sides by $$\exp \left[ \int^t -a(s) \; \mathrm{d}s + k_1 \right] $$, we get
 * $$ x(t) = exp \left[ \int^{t_0} -a(s) \; \mathrm{d}s - \int^{t} \{-a(s)\} \; \mathrm{d}s \right] x(t_0) + \int_{t_0}^t \exp \left[ \int^\tau -a(s) \; \mathrm{d}s -  \int^{t} \{-a(s)\} \; \mathrm{d}s  \right]  b(\tau) u(\tau) \mathrm{d} \tau $$

$$\Rightarrow x(t) = exp \left[ \int^t_{t_0} a(s) \; \mathrm{d}s\right] x(t_0) + \int_{t_0}^t \exp \left[ \int^t_\tau a(s) \; \mathrm{d}s   \right]  b(\tau) u(\tau) \mathrm{d} \tau \qquad\blacksquare$$ Pavel Bhowmik (talk) 14:14, 1 October 2012 (UTC)

=Problem 3.11* - Free Vibration of Coupled Pendulums =

Given:
Pendulums: $$a=0.3,l=1,k=0.2,m_{1}g=3,m_{2}g=6$$ No applied forces Initial Conditions: $$\theta_{1}=+1,\dot{\theta}_{1}(0)=-2,\theta_{2}(0)=-1/2,\dot{\theta}_{2}(0)=+1$$

Find:
1. Use matlab's ode45 command to integrate system 2. Use $$x(t)=[exp{A(t-t_{0})}]*x(t_{0})+\int_{t_{0}}^{t}[exp{A(t-\tau)}]Bu(\tau)d\tau$$ to find the solution from $$t=[0,7]$$ 3. Plot results from 1 & 2

Solution:
Part 1:

The formulas for terms a-f are given in the matlab code

Part 2:

Part 3:

=Problem 3.12* Derive the 2nd exactness condition to prove that a differential equation is exact =

Given:
1st relation of the 2nd exactness condition

2nd relation of the 2nd exactness condition

Find:
Part 1. Derive the exactness condition with respect to p.

Solution:
First we obtain equation 12.1 in terms of p using equation 12.2, <br\>
 * $$g(x,y,p)=\phi_{x}+\phi_{y}p$$

<br\> Then we differentiate with respect to p,


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

<br\> Collecting terms we find that, <br\>
 * $$\frac{\partial}{\partial p}g(x,y,p)=\frac{\partial}{\partial p}\phi_{x}+2\frac{\partial}{\partial y}p\phi_{y}+\frac{\partial}{\partial p}p\phi_{y}$$

<br\> This notation can be simplified to the final solution for the 2nd relation of the 2nd exactness condition, <br\>
 * $$g_{pp}=f_{xp}+pf_{yp}+2f_{y}$$

<br\>

Find:
Part 2. Equation 12.4 is grouped in the correct format to pass the 1st exactness condition so show that it passes the 2nd exactness condition. Show that equation 12.4 is an exact differential equation.

Solution:
Find g(x,y,p) and f(x,y,p) for the given equation, <br\>
 * $$g(x,y,p)=x(p)^2+yp$$

<br\>
 * $$f(x,y,p)=xy$$

<br\> The derivative terms for the 1st relation in equation 11.5 are as follows, <br\>
 * $$f_{xx}=0, f_{xy}=1, f_{yy}=0, g_{xp}=2p, g_{yp}=1, g_{y}=p$$

<br\> Plugging these terms into equation 11.5 yields,' <br\>
 * $$0+2p(1)+(0)p^2=2p+1(p)-p$$

<br\> Simplifying and canceling terms gives, <br\>
 * $$2p=2p$$

<br\> So the equation passes the 1st relation of the 2nd exactness equation. Using equation 12.6, the 2nd relation of the 2nd exactness, we can find out if the differential equation is in fact exact. <br\>
 * $$g_{pp}=f_{xp}+pf_{yp}+2f_{y}$$

<br\> where are derivative terms are calculated as follows, <br\>
 * $$g_{pp}=2x, f_{xp}=0, f_{yp}=0, f_{y}=x$$

<br\> Plugging the terms into equation 11.6, <br\>
 * $$2x=0+p(0)+2x$$

<br\> Simplifying we find that both sides are equal to 2x and thus the differential equation is exact, <br\>
 * $$2x=2x$$

<br\> =Contribution Table=

= References =