User:Egm6321.f11.team4/HW2

=Problem 2.1-Verify the Given Equation=

Given
The Legendre differential equation:

When $$\displaystyle n=1 $$, it becomes:

The two homogeneous solutions for equation (1.2) are:

Find
Verify that when $$\displaystyle n=1 $$,

is true.

Solution
Substituting $$\displaystyle y $$ for $$\displaystyle x $$ in equation (1.2):

Substituting $$\displaystyle y $$ for $$\displaystyle \frac{x}{2}ln(\frac{1+x}{1-x}) $$ in equation (1.2):

Derive $$\displaystyle y^2_H = \frac{x}{2}ln(\frac{1+x}{1-x}) $$ to get its first and second order derivitives:

Substituting in equation (1.7):

So equation (1.5)is true.

Author
Contributed by Yu Chen

=Problem 2.2 - Verify the Solution of an Equation=

Given
A differential equation:

Find
is the solution to equation (2.1).

Solution
Substituting $$\displaystyle p' $$ for (2.3) in equation (2.1):

End of proof.

Author
Contributed by Yu Chen

= Problem 2.3 = From [[media:Pea1.f11.mtg7.djvu|Mtg 7-6]]

Given
The Eq. 2 in the class note [[media:Pea1.f11.mtg7.djvu|7-6]]. The Eq. 1 in the class note [[media:Pea1.f11.mtg7.djvu|7-6]].

Find
Show that the Eq. (3.1) is linear in $$\displaystyle y^{\prime}$$, and that the Eq. (3.1) is in general an N1-ODE. But the Eq. (3-1) is not the most general N1-ODE as represented by the Eq. (3.2). Give an example of a more general N1-ODE.

Show that the Eq. (3.1) is linear in the first derivative of y.
Divide the Eq. (3.1) by $$\displaystyle N(x,y) $$.

Then, the following is true. - The Eq. (3.5) proves that the Eq. (3.1) is linear in $$\displaystyle y^{\prime} $$.

Show that the Eq. (3.1) is N1-ODE.
- The Eq. (3.1) is the 1st order differential equation since the highest order of derivative is 1. - The Eq. (3.1) is Ordinary Differential equation because the differential equation includes one dependent variable and its derivatives.

To see whether the Eq. (3.1) is the linear differential equation, the superposition requirement on the Eq. (3.4) is checked. In other words, the Eq. (3.5) has to be true for the equation to be linear.

Since the RHS of the Eq. (3.6) is not equal to the RHS of the Eq. (3.7), the Eq. (3.1) does not satisfy the superposition condition.

- Since the Eq. (3.1) does not satisfy the superposition, the Eq. (3.1) is nonlinear differential equation.

The Eq. (3.1) is not the most general N1-ODE.
The Eq. (3.1) becomes the 1st order linear ordinary differential equation in particular case when the functions $$\displaystyle M(x,y)$$, $$\displaystyle N(x,y)$$ are in the form of the following.

Since the RHS of the Eq. (3.10) is equal to the RHS of the Eq. (3.11), the Eq. (3.1) satisfies the superposition condition in this particular case. - Therefore, the Eq. (3.1) is not the most general 1st order nonlinear differential equation.

Give an example of a more general N1-ODE.
An example of a general the 1st order nonlinear ordinary differential equation which cannot be converted into the particular form in the Eq. (3.1) is the following. where The Eq. (3.13) can be expressed more generally as following.

where $$\displaystyle F \left( y^{\prime} \right) $$ is any function of $$\displaystyle y^{\prime} $$ (nonlinear in general). If the function $$\displaystyle F(\cdot) $$ has no explicit inverse $$\displaystyle F^{-1}(\cdot) $$, then the Eq. (3.15) cannot be converted into the Eq. (3.1).

Author
Contributed by Shin

=''' Problem 2.4 - Show that the Eq. (3) in page 8-1 is a N1-ODE. '''= From [[media:Pea1.f11.mtg8.djvu|Mtg 8-1]]

Given
The Eq. 3 in the class note [[media:Pea1.f11.mtg8.djvu|8-1]].

Find
Show the Eq. (4.1) is non-linear 1st order ODE.

Solution
- The Eq. (4.1) is 1st order equation since the highest order of derivative is 1 as shown in the Eq. (4.2). - The Eq. (4.1) is Ordinary Differential equation because the differential equation includes one dependent variable and its derivatives. To see whether the Eq. (4.1) is the linear differential equation, the superposition requirement on the Eq. (4.3) is checked. In other words, the Eq. (4.4) has to be true for the equation to be linear.

Since the RHS of the Eq. (4.5) is not equal to the RHS of the Eq. (4.6), the Eq. (4.1) does not satisfy the superposition condition. - Since the Eq. (4.1) does not satisfy the superposition, the Eq. (4.1) is nonlinear differential equation.

Author
Contributed by Shin

=''' Problem 2.5 - Explain the Eq. (4) in p.8-1 and the Eq. (1) in p.8-2 '''= From [[media:Pea1.f11.mtg8.djvu|Mtg 8-2]]

Given
The Eq. (4) in the class note [[media:Pea1.f11.mtg8.djvu|8-1]] and the Eq. (1) in the class note [[media:Pea1.f11.mtg8.djvu|8-2]].

The Eq. (1) in the class note [[media:Pea1.f11.mtg8.djvu|7-6]] and the Eq. (2) in the class note [[media:Pea1.f11.mtg8.djvu|7-6]].

Find
Explain the Eq. (5.1) satifies the the Eq. (5.3) but can be converted to the Eq. (5.4) with the definition in the Eq. (5.2).

Solution
Move the first term in the LHS of the Eq. (5.1) to the LHS. Take the cube root on both side. Move back the term in the RHS to the LHS. Then, the Eq. (5.7) can be converted into the form of the Eq. (5.4) as shown below. when

Author
Contributed by Shin

=Problem 2.6-Show Linearly Independence= From [[media:Pea1.f11.mtg9.djvu|Mtg 9-1]]

Given
The two homogeneous solutions for equation (1.2) in problem 1.

Find
Show that the two solutions a linearly independent, and plot them.

Plotting
To make sense $$\displaystyle x $$ must be defined on the interval $$\displaystyle (-1,1) $$.

Plot $$\displaystyle y_{H}^{1} $$ and $$\displaystyle y_{H}^{2} $$. Matlab Code:


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Proof of Linearly Independence
To prove linearly independence of $$\displaystyle y_{H}^{1} $$ and $$\displaystyle y_{H}^{2} $$ is to show that $$\displaystyle \forall \alpha \in \mathbb R, y_{H}^{1} \ne \alpha y_{H}^{2}$$.

That is $$\displaystyle \exists \hat x $$ such that $$\displaystyle y_{H}^{1}(\hat x) \ne \alpha y_{H}^{2}(\hat x) $$.

Take $$\displaystyle \alpha = 0 $$ for example.When $$\displaystyle \alpha = 0 $$, $$\displaystyle y_{H}^{1} = \alpha y_{H}^{2} $$ yields $$\displaystyle x=0 $$.

But when $$\displaystyle x=\frac{1}{2} $$, $$\displaystyle y_{H}^{1} \ne \alpha y_{H}^{2} $$.

So $$\displaystyle \forall \alpha \in \mathbb R, y_{H}^{1} \ne \alpha y_{H}^{2}$$.

Author
Contributed by Yu Chen

= Problem 2.7 - Find and Verify the N1-ODE = From [[media:Pea1.f11.mtg9.djvu|Mtg 9-2]]

Find
And show that is an N1-ODE.

Solution
According to the particular class of N1-ODEs:
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$$ M(x,y)+N(x,y){y}'=0 \displaystyle $$ (7.3) We have
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$$ M=\frac{\partial \phi }{\partial x}=2xy^{3/2}+\frac{3}{xlog10}\displaystyle $$ (7.4) and
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$$ N=\frac{\partial \phi }{\partial y}=\frac{3}{2}x^{2}y^{1/2}+\frac{2}{ylog10}\displaystyle $$ (7.5)
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Plugging into the equation (7.3) we get
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$$ G(x,y,{y}')=(2xy^{3/2}+\frac{3}{xlog10})+(\frac{3}{2}x^{2}y^{1/2}+\frac{2}{ylog10}){y}' \displaystyle $$ (7.6)
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Obviously, we have already known that So the equation is an N1-ODE.

To be specify, an exact N1-ODE must satisfy the following 2 conditions :

Condition 1
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$$ M(x,y)+N(x,y){y}'=0 \displaystyle $$ (7.7) Condition 2
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$$ M_{y}(x,y)=0=N_{x}(x,y) \displaystyle $$ (7.8)
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For condition 1: We have proved that equation can be put in the particular form.

For Condition 2:
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$$ \frac{\partial M(x,y)}{\partial y}=3xy^{1/2} \displaystyle $$ (7.9)
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$$ \frac{\partial N(x,y)}{\partial x}=3xy^{1/2} \displaystyle $$ (7.10) Therefore
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$$ \frac{\partial M(x,y)}{\partial y}=\frac{\partial N(x,y)}{\partial x}=3xy^{1/2} \displaystyle $$ (7.11)
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In conclusion, $$ G(x,y,{y}') \displaystyle $$ is an exact N1-ODE.

Author
Contributed by Kexin Ren

= Problem 2.8 - 1st Exactness Condition = From [[media:Pea1.f11.mtg9.djvu|Mtg 9-3]]

Given
An N1-ODE as following

Find
Does the N1-ODE satisfy the first exactness condition

Solution
Express the N1-ODE above more generally as $$\displaystyle \bar M(x,y) + \bar N(x,y)\ F(y') = 0 $$ in order to find whether it can be put in the particular form.

Rearrange the equation(8.1), we have We can easily figure out that $$\displaystyle F(y')$$ has no explicit inverse $$\displaystyle F^{-1}(y')$$, so it can not be put in the particular form.

Therefore, the N1-ODE(8.1) does not satisfy the first exactness condition.

Author
Contributed by Kexin Ren

= Problem 2.9 - 2nd Exactness Condition = From [[media:Pea1.f11.mtg9.djvu|Mtg 9-4]]

Given

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$$ M=\frac{\partial \phi(x,y)}{\partial x}\displaystyle $$ (9.1)
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$$ N=\frac{\partial \phi(x,y) }{\partial y}\displaystyle $$ (9.2)
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$$ \frac{\partial M(x,y)}{\partial y}=\frac{\partial N(x,y)}{\partial x} \displaystyle $$ (9.3)
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Find
Review calculus, and find the minimum degree of differentiability of the function $$\displaystyle \phi(x,y) $$. State the full theorem and provide a proof.

Solution
Before we solute the problem, we remind the following theorems first

Mixed Derivative Theorem
If $$\displaystyle f(x,y)$$ and its partial derivatives $$\displaystyle f_{x}, f_{y}, f_{xy}$$ and $$\displaystyle f_{yx}$$ are defined in a neighborhood of $$\displaystyle (a,b)$$ and all are continuous at $$\displaystyle (a,b)$$, then $$\displaystyle f_{xy}(a,b)=f_{yx}(a,b)$$.

Mean Value Theorem
Suppose $$\displaystyle f:\mathbb {R}^2\rightarrow \mathbb {R}$$ is differentiable. Let $$\displaystyle X_{0} = (x_{0}, y_{0})$$ and $$\displaystyle X = (x_{0} + h, y_{0} + k)$$.

Then there exists $$\displaystyle C$$ which lies on the line joining $$\displaystyle X_{0}$$ and $$\displaystyle X$$ such that
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$$f(X) = f(X_{0}) + f'(C)(X-X_{0}) \displaystyle $$ i.e, there exists $$ c\in (0,1) \displaystyle $$ such that
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$$f(x_{0} + h, y_{0} + k) = f(x_{0}, y_{0}) + hf_{x}(C) + kf_{y}(C) \displaystyle $$ where $$\displaystyle C = (x_{0} + ch, y_{0} + ck)$$
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Proof
Suppose $$\displaystyle f:\mathbb [a,b]\rightarrow \mathbb R$$
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$$F_{1} = f(a+h,b+k) - f(a+h,b)-f(a,b+k)+f(a,b) \displaystyle $$ (9.4) and
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$$F_{2} = f(x,y+k) - f(x,y) \displaystyle $$ (9.5) According to MVT, we can know
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$$\displaystyle \frac{F_{2}(a+h,b)-F_{2}(a,b)}{h}=\frac{\partial F_{2}}{\partial x}(a+ch,b) $$, where $$ c\in (0,1) \displaystyle $$ (9.6) Because
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$$\displaystyle {F_{2}(a+h,b)-F_{2}(a,b)}=f(a+h,b+k) - f(a+h,b)-f(a,b+k)+f(a,b)=F_{1} \displaystyle $$ (9.7) We can simply put equation(9.6) into that
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$$\displaystyle \frac{F_{1}}{h}=\frac{\partial F_{2}}{\partial x}(a+ch,b) $$ (9.8) Then focus on the right side of equation(9.8), we can get
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$$\displaystyle \frac{\partial F_{2}}{\partial x}(a+ch,b)=\frac{\partial f}{\partial x}(a+ch,b+k)-\frac{\partial f}{\partial x}(a+ch,b) $$ (9.9) According to MVT again, we have
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$$\displaystyle \frac{\frac{\partial f}{\partial x}(a+ch,b+k)-\frac{\partial f}{\partial x}(a+ch,b)}{k}=\frac{\partial^{2}}{\partial x\partial y}f(a+ch,b+ck) $$ (9.10) Back to equation(9.8),we have
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$$\displaystyle \frac{F_{1}}{hk}=\frac{\partial^{2}}{\partial x\partial y}f(a+ch,b+ck) $$ (9.11) Thus, when $$\displaystyle h\rightarrow 0$$, $$\displaystyle k\rightarrow 0$$, the limit
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$$\displaystyle \lim_{h,k\rightarrow 0}\frac{F_{1}}{hk}=\frac{\partial^{2}f}{\partial x\partial y}(a,b) $$ (9.12) The process will be totally same when we exchange the derivative order to finally get
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$$\displaystyle \lim_{h,k\rightarrow 0}\frac{F_{1}}{kh}=\frac{\partial^{2}f}{\partial y \partial x}(a,b) $$ (9.13) Since $$\displaystyle [a,b]\rightarrow \mathbb {R}$$, we have
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$$\displaystyle \frac{\partial^{2}f}{\partial x \partial y}(x,y)=\frac{\partial^{2}f}{\partial y \partial x}(x,y) $$
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Author
Contributed by Kexin Ren

= Problem 2.10 - Verify thath (1) p.10-4 is indeed the solution for the N1-ODE (2) p.8-6 = From [[media:Pea1.f11.mtg10.djvu|Mtg 10-5]]

Find
Eq. (10.1) is the indeed solution for the N1-ODE (10.2)

Solution
Input Eq. (10.3) to (10.2)

As a result, (10.1) is the indeed solution for (10.2)

Author
Contributed by Chung

= Problem 2.11 - Explain why solving intergrating factor h(x,y) is usually not easy = From [[media:Pea1.f11.mtg11.djvu|Mtg 11-1]]

Find
Explain why solving (11.1) for the integrating factor h(x,y) is usually not easy.

Solution
To solve (11.1), you have to solve partial equation for $$ h_x, h_y$$ and $$M_y,  N_x$$.

Then solve intergral for exponential is not easy to calculate.

So, it is not easy to solve.

Author
Contributed by Chung

= Problem 2.12 - Find h = From [[media:Pea1.f11.mtg11.djvu|Mtg 11-3]]

Find
Find h usin (12.1)

Author
 Contributed by Chung 

= Problem 2.13 - Show (3) p.11-5 = From [[media:Pea1.f11.mtg11.djvu|Mtg 11-5]]

Find
Show that

Solution
for

if c=1; h=x

Input (13.5) into (13.1)

As a result, (13.5) is a solution.

Author
 Contributed by Chung 

= Problem 2.14 - Solve the general L1-ODE-VC =

From [[media:Pea1.f11.mtg6.djvu|Mtg 11-5]]

From [[media:Pea1.f11.mtg6.djvu|Mtg 12-1]]

Given
The general structure of Linear 1st-Order ODEs with Varying Coefficients (L1-ODE-VC) is

$$\displaystyle a_1(x)y'+a_0(x)y=b(x). $$

Find
The solution for $$\displaystyle y $$ if:

$$\displaystyle 1) a_1(x)=1, a_0(x)=x, b(x)=2x+3$$.

$$\displaystyle 2) $$ In general for $$\displaystyle a_1(x), a_0(x), b(x)$$.

$$\displaystyle 3) a_1(x)=x^2+1, a_0(x)=x, b(x)=2x$$.

Solution
$$\displaystyle 1) $$ If $$\displaystyle a_1(x)=1, a_0(x)=x, b(x)=2x+3$$ then $$\displaystyle y'+xy=2x+3$$ then

The integrating factor is $$\displaystyle h= \exp \int^{x}{a_0(x)ds}$$ which is $$\displaystyle h= \exp \int^{x}{sds}$$.

So, the integrating factor is $$\displaystyle h=exp\frac{x^2}{2}$$.

Using the integrating factor and eq. (1) 11-5 yields

$$ y=\exp\frac{-x^2}{2}\int^{x}{\exp\frac{-x^2}{2}*(2s+3)}ds$$.

Solving for $$y$$ yields

$$ y=\exp\frac{-x^2}{2}[2\int^{x}\exp\frac{s^2}{2}sds+3\int^{x}\exp\frac{s^2}{2}ds]$$

$$ y=2x+\exp\frac{-x^2}{2}[3\int^{x}\exp\frac{s^2}{2}ds]+C$$

$$\displaystyle 2) $$ In general for $$\displaystyle a_1(x), a_0(x), b(x)$$.

The integrating factor is $$\displaystyle h=\exp[\int^{x}\frac{a_0(s)}{a_1(s)}ds]$$.

Using eq. (1) 11-5 and replacing the integrating factor and right side function with the general terms, yields

$$\displaystyle y(x)=\frac{1}{\exp[\int^{x}{\frac{a_0(s)}{a_1(s)}}ds]}*\int^{x}\exp[\int^{x}{\frac{a_0(s)}{a_1(s)}ds}]*\frac{b(s)}{a_1(s)}ds$$.

$$\displaystyle3)$$ If $$\displaystyle a_1(x)=x^2+1, a_0(x)=x, b(x)=2x$$ then

The integrating factor is

$$\displaystyle h=\exp[\int^{x}\frac{s}{s^2+1}ds]$$.

This simplifies to

$$\displaystyle h=\exp*\frac{1}{2}\ln(x^2+1)=(x^2+1)^{1/2}$$.

So, the L1-ODE-VC becomes

$$\displaystyle (x^2+1)^{1/2}y'+(x^2+1)^{1/2}*\frac{x}{x^2+1}y=(x^2+1)^{1/2}*\frac{2x}{x^2+1}$$.

Simplifying reduces it to

$$\displaystyle (x^2+1)^{1/2}\frac{dy}{dx}+\frac{x}{(x^2+1)^\frac{1}{2}}y=\frac{2x}{(x^2+1)^\frac{1}{2}}$$.

Rearranging yields

$$\displaystyle (x^2+1)^{1/2}dy=\frac{x(2-y)}{(x^2+1)^\frac{1}{2}}dx$$.

Differenting both sides yields

$$\displaystyle (x^2+1)^\frac{1}{2}y=(2-y)*(x^2+1)^\frac{1}{2}$$.

Multiply through by $$\displaystyle (x^2+1)^\frac{1}{2}$$ and rearrange yields

$$\displaystyle y=\frac{C}{(x^2+1)^\frac{1}{2}}$$.

Author
Contributed by Allen

= Problem 2.15 - Show That K1 In (3) p.11-4 Is NOT Necessary = From [[media:Pea1.f11.mtg12.djvu|Mtg 12-2]]

Given
Eq. (3) p.11-3 is

$$\displaystyle y'+a_0(x)y=b(x)$$.

Eq. (3) p.11-4 is

$$\displaystyle h(x)=\exp[\int^x{a_0(s)ds+k_1}]$$.

Find
Show that the integration constant $$\displaystyle k_1$$ in (3) p.11-4 is NOT necessary.

Solution
Multiply (3) p.11-3 by the integrating factor, (3) p.11-4, which yields the following:

$$\displaystyle \exp[\int^x{a_0(s)ds+k_1}]*y'+\exp[\int^x{a_0(s)ds+k_1}]*a_0(x)y=\exp[\int^x{a_0(s)ds+k_1}]*b(x)$$.

Next, seperate $$\displaystyle \exp[\int^x{a_0(s)ds+k_1}]$$ into $$\displaystyle \exp[\int^x{a_0(s)ds}]*\exp[k_1]$$, which yields the following:

$$\displaystyle \exp[k_1]*\exp[\int^x{a_0(s)ds}]*y'+\exp[k_1]*\exp[\int^x{a_0(s)ds}]*a_0(x)y=\exp[k_1]*\exp[\int^x{a_0(s)ds}]*b(x)$$.

$$\displaystyle \exp[k_1]$$ cancels out of the equation showing $$\displaystyle k_1$$ is NOT necessary.

Author
Contributed by Allen

= Problem 2.16 - Show Solution from the Lecture Agrees with King 2003 = From [[media:Pea1.f11.mtg12.djvu|Mtg 12-2]]

Given
Eq. (3) p.11-3 is

$$\displaystyle y'+a_0(x)y=b(x)$$.

Eq. (3) p.11-4 is

$$\displaystyle h(x)=\exp[\int^x{a_0(s)ds+k_1}]$$.

Eq. (1) p.11-5 is

$$\displaystyle y(x)=\frac{1}{h(x)}[\int^x[h(s)b(s)ds+k_2]$$.

Find
Show that the solution of (3) p.11-3 in (1) p.11-5 agrees with the result presented in King 2003 p.512.

$$\displaystyle y(x)=Ay_H(x)+y_P(x)$$

Solution
King defines the L1-ODE-VC as

$$\displaystyle y'+P(x)y=Q(x)$$.

King also defines the integrating factor as

$$\displaystyle h(x)=\exp[\int^x{P(t)dt}]$$ where t is a dummy variable.

The equation $$\displaystyle y(x)=Ay_H(x)+y_P(x)$$ is defined with

$$\displaystyle y_H(x)=A*\exp[-\int^x{P(t)}dt]$$

$$\displaystyle y_P(x)=\exp[-\int^x{P(t)dt}]\int^x{Q(s)}\exp[\int^x{P(t)dt}]ds$$

and $$\displaystyle A$$ is constant of integration.

Expand Equation (1) P.11-5 with Equations (3) p.11-3 and (3) p.11-4 which yields

$$\displaystyle y(x)=\frac{1}{\exp[\int^x{a_0(s)ds}]}*[\int^x{[\exp[\int^x{a_0(s)ds}]*b(s)ds]+k_2}]$$

Further expand the above equation and you can break up $$\displaystyle y_H(x)$$ as

$$\displaystyle y_H(x)=\frac{k_2}{\exp[\int^x{a_0(s)ds}]}$$.

This shows that $$\displaystyle k_2=A$$ and $$\displaystyle a_0(s)=P(t)$$.

Also, it can be shown that $$\displaystyle b(s)=Q(t)$$ from the equation below:

$$\displaystyle y_P(x)=\frac{1}{\exp[\int^x{a_0(s)ds}]}*[\int^x{[\exp[\int^x{a_0(s)ds}]*b(s)ds]}]$$.

Author
Contributed by Allen

=Problem 2.17 - Homogenous and Particular solutions of general Non-Homogenous L1-ODE-VC= From [[media:Pea1.f11.mtg12.djvu|Mtg 12-2]]

Given
A General non-homogenous L1-ODE-VC (Linear, 1st order, Ordinary Differential Equation with Variable Coefficient),

$$\displaystyle P(x)\,y' + Q(x)\,y = \underbrace{R(x)}_{\color{blue}{\neq 0}}$$<p style="text-align:right"> (17.1)

which can be written in the form given below,

$$\underbrace{\color{blue}{1}}_{\color{blue}{a_1(x)}} \cdot \underbrace{y'}_{\color{blue}{y^{(1)}}} + \underbrace{\frac{Q(x)}{P(x)}}_{\color{blue}{a_0(x)}}y^{\color{blue}{(0)}} = \underbrace{\frac{R(x)}{P(x)}}_{\color{blue}{b(x)}} $$ <p style="text-align:right"> (17.2)

The form can be written as,

$$\displaystyle y' + a_0(x)y^{(0)} = b(x) $$ <p style="text-align:right"> (17.3a)

Find
Find the homogenous counterpart, $$\displaystyle y_H(x) $$ of Eq. (18-3) above, by solving,

$$\displaystyle y' + a_0(x)y = 0 $$<p style="text-align:right"> (17.3b)

and also find the particular solution, $$\displaystyle y_P(x) $$ by use of the method of Variation of Parameters

Solution
Eq. 17.3b can be rearranged to,

$$\displaystyle \frac{y'}{y} = - a_0(x) $$<p style="text-align:right"> (17.4)

Integrate both sides, we get,

$$\displaystyle \log y = - \int^x a_0(s)ds + K_1 $$<p style="text-align:right"> (17.5)

Thus homogenous solution, $$\displaystyle y_H (x) $$, can be written as,

$$\displaystyle A.y_H (x) = \exp [- \int^x a_0(s)ds ]. \underbrace{k_1}_{A} $$<p style="text-align:right"> (17.6)

Thus, $$\displaystyle y_H (x) $$ is

$$\displaystyle y_H (x) = \exp [- \int^x a_0(s)ds ] $$<p style="text-align:right"> (17.7)

For particular solution, $$\displaystyle y_P (x) $$ we use Variation of Parameters, by which we require y_P(x) to be following,

$$\displaystyle y_p(x) = \sum_{i=1}^{n} c_i(x) y_i(x) $$ <p style="text-align:right"> (17.8)

for $$ n^{th} $$ order non-linear homogenous ODE. For our case, n = 1, therefore we get,

$$\displaystyle y_p(x) = c_1(x) y_H(x) $$ <p style="text-align:right"> (17.9)

It is important to note for method of Variation of Parameters that above, $$\displaystyle c_1 $$ is a function of $$\displaystyle x $$, and is determined by substituting the particular solution, i.e. Eq. 17.9 into Eq. 17.3b, the ODE of interest.

Thus,

$$\displaystyle y'_H(x) + y'_P(x) + a_0(x)[y_H(x) + y_P(x)] = b(x)$$ <p style="text-align:right"> (17.10)

Substitute Eq. 17.9 into Eq. 17.10 to get,

$$\displaystyle y'_H(x) + [c'_1(x)y_H(x) + c_1(x)y'_H(x)] + a_0(x)y_H(x) + a_0(x)[y_H(x) + c_1(x)y_H(x)] = b(x)$$ <p style="text-align:right"> (17.11)

and, rearrange the terms,

$$\displaystyle \underbrace{[y'_H(x) + a_0(x)y_H(x)]}_{=0} + c'_1(x)y_H(x) + c_1(x)\underbrace{[y'_H(x) + a_0(x)y_H(x)]}_{=0} = b(x)$$<p style="text-align:right"> (17.12)

Therefore $$\displaystyle c_1(x) $$ is given by,

$$\displaystyle c'_1(x) = \frac{b(x)}{y_H(x)}$$<p style="text-align:right"> (17.13)

Integrate both sides w.r.t. x,

$$\displaystyle c_1(x) = \int^x \frac{b(s)}{y_H(s)} ds$$<p style="text-align:right"> (17.14)

Substituting for Eq. 17.7 for expression of $$\displaystyle y_H(x) $$ into above, we will get,

$$\displaystyle c_1(x) = \int^x \underbrace{\exp [ \int^s a_0(t)dt ]}_{=h(s)} b(s) ds$$<p style="text-align:right"> (17.15)

Therefore particular solution can finally be written as,

$$\displaystyle y_p(x) = [\int^x h(s) b(s) ds] \frac{1}{h(s)}$$<p style="text-align:right"> (17.16)

Author
Contributed by Ankush

= Problem 2.18 - IFM for L1-ODE-VC = From [[media:Pea1.f11.mtg12.djvu|Mtg 12-5]]

Given
A particular L1-ODE-VC, (4) p.12-4

$$\displaystyle [x^4y + 10] + (\frac{1}{2} x^2)\,y'= 0 $$ <p style="text-align:right"> (18.1)

Find
Find if Eq. 18.1 is exact? If not, then find the integrating factor h(x) to make it exact.

Solution
Above equation, has the form as in (2), p.7-6,

$$\displaystyle \underbrace{[x^4y + 10]}_{M(x,y)} + \underbrace{(\frac{1}{2} x^2)}_{N(x,y)}\,y'= 0 $$ <p style="text-align:right"> (18.2)

And, above equation is linear in $$\displaystyle y' $$ and N1-ODE (Non-Linear 1st order Ordinary Differential Equation).

To check, if it is exact, we check 2nd condition for exactness, (3), p.9-3, i.e.,

$$\displaystyle M_y(x,y) = N_x(x,y) $$

Substituting the expression for $$\displaystyle M(x,y) $$ and $$\displaystyle N(x,y) $$ as shown in Eq. 18.2 into above equation,

$$\displaystyle \frac{\partial}{\partial x} (x^4y + 10)  =? \ne \frac{\partial}{\partial x} (\frac{x^2}{2}) $$ <p style="text-align:right"> (18.3)

Where, above $$\displaystyle =?\ne $$ denotes that it can be either $$\displaystyle = $$ or $$\displaystyle \ne $$, Upon further differentiation we get,

$$\displaystyle x^4 \ne x \forall x $$

Therefore the given equation is not exact, and we use Euler Integrating Factor Method (IFM) to find h, integrating factor to make the given equation exact. For this we notice that,

$$\displaystyle - \frac{1}{N}(N_x - M_y) = - \frac{1}{\frac{x^2}{2}}(x-x^4) = (-\frac{2}{x} + 2x^2) \rightarrow n(x) $$ <p style="text-align:right"> (18.4)

Above given expression is thus a function of $$\displaystyle x $$ only, (2) p.11-2 and this means that h is a function of x only. Therefore we can integrate,

$$\displaystyle \frac{h_x}{h} = (- \frac{2}{x} + 2x^2) dx $$ <p style="text-align:right"> (18.5)

to get,

$$\displaystyle \int^x \frac{h_x}{h} = \int^x (- \frac{2}{x} + 2x^2) dx $$ <p style="text-align:right"> (18.6)

From above equation, we can solve for $$\displaystyle h $$,

$$\displaystyle h(x) = exp[\int^x (- \frac{2}{x} + 2x^2) dx + k] $$ <p style="text-align:right"> (18.7)

where, k is integration constant. Above expression can be simplified in following steps to get the final expression of integration factor, h which will make the given L1-ODE-VC Eq. 18.1 exact,

$$\displaystyle h(x) = exp[-2\log(x) + \frac{2}{3}x^3] $$

$$\displaystyle h(x) = exp(-2\log(x)).exp(\frac{2}{3}x^3) $$

And, thus the final form of $$\displaystyle h(x) $$, the integrating factor is,

$$\displaystyle h(x) = exp(\frac{2}{3}x^3).\frac{1}{x^2} $$ <p style="text-align:right"> (18.8)

By multiplying both sides of Eq. 18.1 by $$\displaystyle h(x) $$, it can be verified that the second condition of exactness, (3)p.9-3 is now satisfied.

Author
Contributed by Ankush

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