User:Egm6321.f10.team4/HW2

= Problem 1 - Linearity test of ODEs  =

From Meeting 7, p 7-1.

Given

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

$$\displaystyle \ {\rm{M}}\left( {x,y} \right) + {\rm{N}}\left( {x,y} \right)\frac = 0 $$ $$
 * $$\displaystyle (Eq. 1-1)
 * }
 * }

Find
Verify the given general form of N1-ODE-VC is non-linear.

Solution
Let (Eq.1-1) be a function of H(y)
 * {| style="width:100%" border="0" align="left"

$$\displaystyle H(y)= {\rm{M}}\left( {x,y} \right) + {\rm{N}}\left( {x,y} \right)\frac $$
 * }
 * }

For the given general equation to be linear, it has to meet both of the following conditions: H is a dummy function and alpha and beta are arbitrary constants.


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

$$\displaystyle H(\alpha )=\alpha H $$ $$
 * $$\displaystyle (Eq. 1-2)
 * }
 * }


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

$$\displaystyle H(\alpha+\beta)=H(\alpha)+H(\beta) $$ $$
 * $$\displaystyle (Eq. 1-3)
 * }
 * }

By the first condition of linearity,
 * {| style="width:100%" border="0" align="left"

$$\displaystyle H(\alpha )= {\rm{M}}\left( {x,\alpha y} \right) + {\rm{N}}\left( {x,\alpha y} \right)(\alpha y)' $$


 * }
 * }


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

$$\displaystyle \alpha H = \alpha[{\rm{M}}\left( {x,y} \right) + {\rm{N}}\left( {x,y} \right)y'] $$ Thus,
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle H(\alpha ) \neq \alpha H' $$
 * }
 * }

By the second condition of linearity,
 * {| style="width:100%" border="0" align="left"

$$\displaystyle H(\alpha+\beta)={\rm{M}}\left( {x,\alpha + \beta} \right) + {\rm{N}}\left( {x, \alpha + \beta} \right) \left(\alpha + \beta \right)' $$
 * }
 * }


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

$$\displaystyle H(\alpha)+H(\beta)={\rm{M}}\left( {x,\alpha} \right) + {\rm{N}}\left( {x, \alpha} \right)(\alpha)'+{\rm{M}}\left( {x,\beta} \right) + {\rm{N}}\left( {x, \beta} \right)(\beta)' $$
 * }
 * }

Thus,
 * {| style="width:100%" border="0" align="left"

$$\displaystyle H(\alpha+\beta) \neq H(\alpha)+H(\beta) $$
 * }
 * }

Since the given equation satisfies neither of the linearity conditions, the given general N1-ODE-VC is proven to be non-linear.

Egm6321.f10.team4.Yoon 00:49, 16 September 2010 (EDT) - Primary Author & Editor

= Problem 2 - Order of a differential equation and test of linearity =

From Meeting 7, p. 7-1.

Given

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

$$ \left[ {4x^1 + \sin (y)} \right] + \left( {x^2 y^3 } \right)y' = 0$$ $$
 * $$\displaystyle (Eq. 2-1)
 * }
 * }

Find
Verify the given equation is an example of a N1-ODE.

Solution
The order of any equation can be easily determined just by inspection. In the case of Eq. (2-1), the highest-order derivative is $$y'$$ and it is of order one. Thus, Eq. (2-1) is a first order differential equation.

To verify that is it a nonlinear equation, the linearity test from the previous problem (Problem 1) should be applied. Let Eq. (2-1) be a function of H(y)
 * {| style="width:100%" border="0" align="left"

$$H(y)= \left( {4x^1 + \sin (y)} \right) + \left( {x^2 y^3 } \right)y' = 0$$ $$
 * $$\displaystyle (Eq. 2-2)
 * }
 * }

Let us now apply Eq. (1-2). The left-hand side will take the following form:


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

H(\alpha y) = 4x^7 + \sin (\alpha y) + (x^2 (\alpha y)^3 )y'$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2-3)
 * }
 * }

while the right-hand side will take this form:


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

\alpha H(y) = \alpha \left( {4x^7 + \sin (y)} \right) + \alpha (x^2 y^3 )y' $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2-4)
 * }
 * }

Comparing Eq. (2-3) and Eq. (2-4), it is clear that Eq. (2-1) is nonlinear.

Egm6321.f10.team4.petralanda.n 21:00, 18 September 2010 (UTC) - Author

= Problem 3 - Graphical interpretations of nonlinear variables  =

From Meeting 8, p. 8-1.

Given
General form of a homogeneous Non-linear second order differential equation(N2_ODE_VC) :


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

$$\,{a}_{2} \left( x\right)\ddot{y}+{a}_{1} \left(x \right) \dot{y}+{a}_{0} \left( x\right)y=f\left(x \right)$$
 * $$\,(Eq.3.1)$$
 * }

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

$${y}^{1}_{H}\left(x \right)=x$$
 * $$\,(Eq.3.2)$$
 * }


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

$${y}^{2}_{H}\left(x \right)= \frac{x}{2}log \left( \frac{1+x}{1-x} \right)-1$$
 * $$\,(Eq.3.3)$$
 * }

Find
Plot $$ y_H ^1 (x) $$ and $$ y_H ^2 (x) $$ and show $$y_H ^1 (x) \ne \alpha y_H ^2 (x)$$.

Where $$\forall x \in \real $$  and   $$\forall \alpha  \in \real$$

Hint: Find $$\hat x$$ such that $$ y^1 _H \left( {\hat x} \right) - \alpha y^2 _H \left( {\hat x} \right) \ne 0$$

Solution
Assume that,


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

$$ \,{y^1_{H}} \left( x\right) = \alpha{y^2_{H}} \left( x\right) $$
 * $$\,(Eq.3.4)$$
 * }

by our assumption, $$\displaystyle \alpha$$ must be linear so that $$\displaystyle \alpha$$ can satisfy the following (eq 3.5) for all $$\displaystyle x$$ in the $$\real$$eal domain


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

$$\,\alpha= \frac{{y^1_{H}}\left( x\right)}{{y^2_{H}} \left( x\right)}= \frac{x}{\frac{x}{2}log \left( \frac{1+x}{1-x} \right)-1}$$


 * $$\,(Eq.3.5)$$
 * }

However, $$\displaystyle \alpha$$ is proven not to be linear by figure.2, our Assumption (Eq. 3.3) is false.

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

$$ \,{y^1_{H}} \left( x\right) \neq \alpha{y^2_{H}} \left( x\right) $$
 * $$\,(Eq.3.6)$$
 * }



Egm6321.f10.team4.Yoon 00:49, 16 September 2010 (EDT) - Primary Author & Editor

68.17.105.125 02:47, 22 September 2010 (UTC)Egm6321.f10.team4.osentowski - Editor

= Problem 4 - Exactness of N1-ODEs =

From Meeting 8, p. 8-2.

Given

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

$$ \varphi (x,y) = x^2 y^{3/2} + \log (x^3 y^2 ) = k$$ $$
 * $$\displaystyle (Eq. 4-1)
 * }
 * }

Find
Part A

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

$$F\left( {x,y,y'} \right) = \frac $$ $$
 * $$\displaystyle (Eq. 4-2)
 * }
 * }

Part B

Verify that $$ \displaystyle F $$ is an exact N1-ODE.

Part C

Create an additional three examples.

Solution
Part A

The total derivative given in Eq. (4-2) can be expanded to take the following form:


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

$$\frac = \frac + \frac \frac $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 4-3)
 * }
 * }

Evaluating the first term in (Eq. 4-3) will yield the following:


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

$$ \frac{\partial } \left( {x^2 y^{3/2} + \log (x^3 y^2 )} \right) = 2xy^{3/2}  + \frac{3} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 4-4)
 * }
 * }

Evaluating the second term in (Eq. 4-3) will yield the following:


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

$$ \frac{\partial } \left( {x^2 y^{3/2} + \log (x^3 y^2 )} \right) = \frac{3} {2}x^2 y^{1/2} + \frac{2} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 4-5)
 * }
 * }

Combining all together we see that there is in fact a function $$F(x,y,y')$$


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

$$ F(x,y,y') = \left( {2xy^{3/2} + \frac{3} } \right) + \left( {\frac{3} {2}x^2 y^{1/2}  + \frac{2} } \right)\frac = 0$$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 4-6)
 * }
 * }

Part B

In order for (Eq. 4-6) to be exact it needs to satisfy the following two conditions:

Condition 1:


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

$$ F(x,y,y') = M(x,y) + N(x,y)\frac = 0 $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 4-7)
 * }
 * }

If we let
 * {| style="width:100%" border="0" align="left"

$$M(x,y) = \left( {2xy^{3/2} + \frac{3} } \right)$$ and  $$N(x,y) = \left( {\frac{3} {2}x^2 y^{1/2}  + \frac{2} } \right)$$
 * }

Then, the first condition is satisfied.

Condition 2:


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

$$ \frac = \frac

$$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 4-8)
 * }
 * }

If we compute the left side of (Eq. 4-8):
 * {| style="width:100%" border="0" align="left"

$$\frac{\partial } \left( {2xy^{3/2} + \frac{3} } \right) = 3xy^{1/2} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 4-9)
 * }
 * }

and if we compute the right side:


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

$$ \frac{\partial } \left( {\frac{3} {2}x^2 y^{1/2} + \frac{2} } \right) = 3xy^{1/2} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 4-10)
 * }
 * }

Comparing (Eq. 4-9) and (Eq. 4-10) it is clear that the second exactness condition is also satisfied. Therefore $$ \displaystyle F(x,y,y') $$ is exact.

Part C

Example 1


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

$$F(x,y,y') = \frac {2}\sin (x) - y\cos (x)=0$$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 4-11)
 * }
 * }

Example 2


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

$$ F(x,y,y') = e^{xy} \left( {\frac{1} {x} + \frac{1} {y}y'} \right)=0 $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 4-12)
 * }
 * }

Example 3


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

$$ F(x,y,y') = \frac + \frac y'=0$$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 4-13)
 * }
 * }

Note: to generate the three examples of Part C, an arbitrary function was assigned to (Eq. 4-8). It was then integrated with respect to x and y to create M and N. Egm6321.f10.team4.petralanda.n 23:11, 18 September 2010 (UTC) - Author

68.17.105.125 02:49, 22 September 2010 (UTC)Egm6321.f10.team4.osentowski - Editor

= Problem 5 - Finding an ODE that is not exact  =

From [http://upload.wikimedia.org/wikiversity/en/1/15/2010_09_09_14_48_41.djvu Mtg. 9], p. 9-2

Given

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

$$ \displaystyle M(x,y) + N(x,y)f(y') = 0 $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 5-1)
 * }
 * }

Find

 * $$\displaystyle f(y')$$ such that there is no analytical solution to $$f(y') = - \frac{M}{N}$$
 * (i.e., such that the N1-ODE cannot be exact).

Solution
Two conditions must be met in order for an equation of the form
 * $$\displaystyle F(x,y,y')=0$$

to be exact:

1) $$\displaystyle F$$ must be in the form:

2) From Eq. (5-2), the partial derivative of $$\displaystyle M(x,y)$$ with respect to $$\displaystyle y$$ must equal the partial derivative of $$\displaystyle N(x,y)$$ with respect to $$\displaystyle x$$:

{| style="width:100%" border="0" align="left"
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle

M_y(x,y) = N_x(x,y) $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 5-3)

For this solution, we need to "invent" some function $$ \displaystyle f(y') $$....

First, suppose {| style="width:100%" border="0" align="left"
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle

f(y')=5y'+7 $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 5-4)

Then: Eq. (5-1) becomes:
 * $$\displaystyle M(x,y) + N(x,y)(5y'+7) = 0$$
 * $$\displaystyle \underbrace{M(x,y) + 7N(x,y)}_{M'(x,y)}+\underbrace{5N(x,y)}_{N'(x,y)}y'= 0$$

This reduces to:
 * $$\displaystyle M'(x,y) + N'(x,y)y' = 0$$

Thus the form of Eq (5-2) is maintained (and we must find another function that will not have an analytical solution).


 * Now, suppose that

{| style="width:100%" border="0" align="left"
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * $$\displaystyle
 * $$\displaystyle

f(y') =\frac{e^{y'}}{sin^3(y')} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 5-5)

Then: Eq. (5-1) becomes:
 * $$\displaystyle M(x,y) + N(x,y)\frac{e^{y'}}{sin^3(y')} = 0$$
 * $$\displaystyle \frac{e^{y'}}{sin^3(y')} = -\frac{M(x,y)}{N(x,y)} $$
 * $$\displaystyle e^{y'} = -\frac{M(x,y)}{N(x,y)}sin^3(y') $$
 * $$\displaystyle y' = -ln \left[\frac{M(x,y)}{N(x,y)}sin^3(y') \right] = ln \left[ N(x,y) \right] -ln \left[ M(x,y) sin^3(y') \right]$$

Rearrange:
 * $$ \displaystyle -ln \left[N(x,y) \right] + ln \left[ M(x,y) sin^3(y') \right]+y'=0$$
 * $$ \displaystyle \underbrace{-ln \left[N(x,y) \right]}_{M'(x,y)} + \underbrace{ln \left[ M(x,y)\right]+ln \left[ sin^3(y') \right]+y'}_{ \neq N'(x,y)y'}=0$$

The equation cannot be arranged in the form of Eq. (5-2), thus Eq (5-5) is an example of a function $$\displaystyle f(y') $$, for which there exists no analytical solution when used in Eq (5-1).


 * Note: There is no need to test the second condition of exactness, Eq (5-3), since we are not given the functions $$\displaystyle M(x,y) $$ and $$\displaystyle N(x,y) $$.

128.227.34.126 14:55, 22 September 2010 (UTC) - Author

= Problem 6 - Verification of the first condition of exactness =

From Meeting 9, p. 9-3.

Given

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

$$\displaystyle y(x) = \sin ^{ - 1} (k - 15x^5 ) $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 6-1)
 * }
 * }


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

$$\displaystyle M(x,y)=75x^4 $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 6-2)
 * }
 * }


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

$$\displaystyle N(x,y)=cos(y) $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 6-3)
 * }
 * }

Find
Verify equation 6-1 satisfies the first condition of exactness,


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

$$\displaystyle M\left(x,y\right)+N(x,y)y'=0 $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 6-4)
 * }
 * }

Solution
Start off by assigning a dummy variable $$\displaystyle u$$


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

$$\displaystyle u = \sin{y} = \sin \left[ \sin ^{ - 1} \left(k - 15x^5 \right) \right] = k - 15x^5 $$
 * }

and


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

$$\displaystyle \frac {du}{dx} = - 75x^4 $$
 * }

Now taking the derivative of $$\displaystyle y$$ and applying the chain rule,


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

$$\displaystyle y' = \frac = \frac \frac = \frac (- 75x^4) $$
 * }

Substituting $$\displaystyle u = \sin{y}$$ back into the equation yields


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

$$\displaystyle y' = \frac $$
 * }

or


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

$$\displaystyle y' = \frac = \frac $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 6-5)
 * }
 * }

Substituting Eq.s (6-2), (6-3), and (6-5) into Eq. (6-4) yields


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

$$\displaystyle M\left(x,y\right)+N(x,y)y'=75x^4+\cos{y} \left( \frac \right) $$
 * }


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

$$\displaystyle =75x^4- 75x^4 = 0 $$
 * }

Thus equation 6-1 meets the first condition of exactness

Egm6321.f10.team4.Auerbach 02:53, 21 September 2010 (UTC) - Author

= Problem 7 - Solving nonhomogenous L1-ODE-VCs =

From Meeting 10 p. 10-3, 10-4, and Meeting 11, p. 11-1.

Given

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

$$
 * $$\displaystyle a_1 (x)y' + a_0 (x)y = b(x)$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 7-1)
 * <p style="text-align:right;">$$\displaystyle (Eq. 7-1)
 * }
 * }

Find
Part A

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

|$$\displaystyle a_0 (x)=x $$ $$
 * $$\displaystyle a_1 (x)=1 $$
 * $$\displaystyle a_1 (x)=1 $$
 * $$\displaystyle b(x)=2x+3$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 7-2)
 * }
 * }

Solve for $$\displaystyle y $$ in Eq. (7-1).

Part B

Assume


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


 * $$\displaystyle a_1 (x) \ne 0$$ $$ \forall {\text{ }}x$$
 * }
 * }
 * }

Then, (Eq. 7-1) becomes


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

y' + \underbrace {\frac }_{P(x)}y = \underbrace {\frac }_{Q(x)} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 7-3)
 * }
 * }

Find a solution for $$\displaystyle y(x) $$ in terms of $$\displaystyle a_0 {\text{, }}a_1 {\text{,}} $$ and $$\displaystyle b. $$

Part C

Give an example where


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

a_1 (x) = x^2 + 1{\text{ }}a_0 (x) = x{\text{ and }}b(x) = 2x$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 7-4)
 * }
 * }

Solution
Part A

Substituting Eq. (7-2) into (7-1),
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle 1y' + xy = 2x + 3$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 7-5)
 * <p style="text-align:right;">$$\displaystyle (Eq. 7-5)
 * }
 * }

Eq. (7-5) can be solved by multiplying both sides of the equation with an integration factor:


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

$$
 * $$\displaystyle h(x)(1y' + xy) = h(x)(2x + 3) $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 7-6)
 * <p style="text-align:right;">$$\displaystyle (Eq. 7-6)
 * }
 * }

where


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


 * $$\displaystyle h(x)=\exp \left[\int^{x} \frac{a_{0}(t)}{a_{1}(t)}dt \right]$$
 * }
 * }
 * }


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


 * $$\displaystyle =\exp \left[\int^{x} \frac{t}{1} dt \right]$$
 * }
 * }
 * }


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


 * $$\displaystyle =\exp \left[\frac{x^2}{2}\right]$$
 * }
 * }
 * }

Rewriting 7-6,


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


 * $$\displaystyle \exp \left[\frac{x^2}{2}\right]y' + \exp \left[\frac{x^2}{2}\right]xy = \exp \left[\frac{x^2}{2}\right](2x + 3) $$
 * }
 * }
 * }

The left side of the equation is equivalent to


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


 * $$\displaystyle \frac{d}{dx}(hy) = hy' + h'y$$
 * }
 * }
 * }

so,


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


 * $$\displaystyle \frac{d}{dx}(hy) = h(2x + 3) $$
 * }
 * }
 * }

integrating both sides


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


 * $$\displaystyle hy = \int^{x} h(s)(2s + 3) ds$$
 * }
 * }
 * }

Dividing both sides of the equation by h,


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


 * $$\displaystyle y = h^{-1} \int^{x} h(s)(2s + 3) ds$$
 * }
 * }
 * }

substituting


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


 * $$\displaystyle y = \exp \left[\frac{-x^2}{2}\right] \int^{x} \exp \left[\frac{s^2}{2}\right](2s + 3) ds$$
 * }
 * }
 * }

completing the integration,


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


 * $$\displaystyle y=\exp \left[\frac{-x^2}{2}\right](2 \exp \left[\frac{x^2}{2}\right]+3 \sqrt{\frac{\pi}{2}} erfi(\frac{x}{\sqrt {2}})+c)$$
 * }
 * }
 * }

expanding,


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


 * $$\displaystyle y=2+3\exp \left[\frac{-x^2}{2}\right] \sqrt{\frac{\pi}{2}} erfi(\frac{x}{\sqrt {2}})+c \exp \left[\frac{-x^2}{2}\right]$$
 * }
 * }
 * }

Part B

starting with equation 7-1,
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle a_1 (x)y' + a_0 (x)y = b(x)$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 7-7)
 * <p style="text-align:right;">$$\displaystyle (Eq. 7-7)
 * }
 * }

Since we have assumed,
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle a_1 (x) \ne 0$$ $$ \forall {\text{ }}x$$
 * }
 * }
 * }

equation 7-7 can be rewritten as


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

$$
 * $$\displaystyle y' + \frac{a_0 (x)}{a_1 (x)}y = \frac{b(x)}{a_1 (x)}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 7-8)
 * <p style="text-align:right;">$$\displaystyle (Eq. 7-8)
 * }
 * }

Equation 7-8 can be solved by multiplying both sides of the equation with an integration factor:


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

$$
 * $$\displaystyle h(x)(y' + \frac{a_0 (x)}{a_1 (x)}y) = h(x)\frac{b(x)}{a_1 (x)} $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 7-9)
 * <p style="text-align:right;">$$\displaystyle (Eq. 7-9)
 * }
 * }

where


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


 * $$\displaystyle h(x)=\exp \left[\int^{x} \frac{a_{0}(t)}{a_{1}(t)}dt\right]$$
 * }
 * }
 * }

Rewriting 7-9,


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


 * $$\displaystyle \exp \left[\int^{x} \frac{a_{0}(t)}{a_{1}(t)}dt\right]y' + \exp \left[\int^{x} \frac{a_{0}(t)}{a_{1}(t)}dt\right]\frac{a_0 (x)}{a_1 (x)}y = \exp \left[\int^{x} \frac{a_{0}(t)}{a_{1}(t)}dt\right]\frac{b(x)}{a_1 (x)} $$
 * }
 * }
 * }

The right side of the equation is equivalent to


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


 * $$\displaystyle \frac{d}{dx}(hy) = hy' + h'y$$
 * }
 * }
 * }

so,


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


 * $$\displaystyle \frac{d}{dx}(hy) = h\frac{b(x)}{a_1 (x)} $$
 * }
 * }
 * }

integrating both sides


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


 * $$\displaystyle hy = \int^{x} h(s)\frac{b(s)}{a_1 (s)} ds$$
 * }
 * }
 * }

Dividing both sides of the equation by h,


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


 * $$\displaystyle y = h^{-1} \int^{x} h(s)\frac{b(s)}{a_1 (s)} ds$$
 * }
 * }
 * }

substituting


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


 * $$\displaystyle y = \exp \left[-\int^{x} \frac{a_{0}(t)}{a_{1}(t)}dt\right] \int^{x} \exp \left[\int^{s} \frac{a_{0}(t)}{a_{1}(t)}dt\right]\frac{b(s)}{a_1 (s)} ds$$
 * }
 * }
 * }

completing the integration,


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

$$
 * $$\displaystyle y = \exp \left[-\int^{x} \frac{a_{0}(t)}{a_{1}(t)}dt\right] \int^{x} \exp \left[\int^{s} \frac{a_{0}(t)}{a_{1}(t)}dt\right]\frac{b(s)}{a_1 (s)} ds + c \exp \left[-\int^{x} \frac{a_{0}(t)}{a_{1}(t)}dt\right]$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 7-10)
 * <p style="text-align:right;">$$\displaystyle (Eq. 7-10)
 * }
 * }

where c is a constant of integration

Part C

Substituting equation 7-4 into equation 7-10 yields


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


 * $$\displaystyle y = \exp \left[-\int^{x} \frac{t}{t^2 + 1}dt\right] \int^{x} \exp \left[\int^{s} \frac{t}{t^2  + 1}dt\right]\frac{2s}{s^2  + 1} ds + c \exp \left[-\int^{x} \frac{t}{t^2  + 1}dt\right]$$
 * }
 * }
 * }

simplifying,


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


 * $$\displaystyle y = \frac{1}{\sqrt{x^2 + 1}} \int^{x} \sqrt{s^2  + 1} \frac{2s}{s^2  + 1} ds + \frac{c}{\sqrt{x^2  + 1}}$$
 * }
 * }
 * }


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


 * $$\displaystyle = \frac{1}{\sqrt{x^2  + 1}} \int^{x}  \frac{2s}{\sqrt{s^2  + 1}} ds + \frac{c}{\sqrt{x^2  + 1}}$$
 * }
 * }
 * }


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


 * $$\displaystyle = \frac{1}{\sqrt{x^2  + 1}} 2 \sqrt{x^2  + 1}+ \frac{c}{\sqrt{x^2  + 1}}$$
 * }
 * }
 * }


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


 * $$\displaystyle y = 2 + \frac{c}{\sqrt{x^2 + 1}}$$
 * }
 * }
 * }

Egm6321.f10.team4.Auerbach 10:55, 22 September 2010 (UTC) - Author

= Solutions Generated by Use of the Integrating Factor Method =

From Meeting 11, p. 11-2.

Given

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

$$\displaystyle \ y\left( x \right) = \frac{1}\int^x h\left( s \right)b\left( s \right)ds = \frac{1}\int h\left( x \right)b\left( x \right)dx+k_2 $$ $$ where
 * <p style="text-align:right;">$$\displaystyle (Eq. 8-1)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \ h(x) = e^{\int^x a_0\left( s \right)b\left( s \right)ds} = e^{\int a_0\left( x \right)b\left( x \right)dx+k_1} $$ $$ where $$\displaystyle k_1 $$ & $$\displaystyle k_2 $$ are integration constants. Also, $$\displaystyle h(x) $$ is a particular case of $$\displaystyle h(x,y) $$, from the equation:
 * <p style="text-align:right;">$$\displaystyle (Eq. 8-2)
 * }
 * }


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

$$\displaystyle \ h(x,y) \left[M(x,y)+N(x,y)y' \right] = 0 $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 8-3)
 * }
 * }

in which case $$\displaystyle h $$ is not a function of $$\displaystyle y $$ (and thus $$\displaystyle \frac{\partial h}{\partial y}= h_y=0 $$).

Find
Part A

Show that $$\displaystyle k_1 $$ is an extraneous constant, when solving by the integrating factor method.

Part B

Show that Eq. (8-1) agrees with King p. 512, where


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

$$
 * $$\displaystyle y(x) = Ay_H (x) + y_P (x)$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 8-4)
 * <p style="text-align:right;">$$\displaystyle (Eq. 8-4)
 * }
 * }

Part C

Find $$ \displaystyle y_H (x)$$ independently.

Hint: solve for


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


 * $$\displaystyle y'+a_0y=0$$
 * }
 * }
 * }

Solution
Part A

In using the Integrating Factor Method (IFM) to find some function $$ h(x,y) $$, such that our solution can be made exact, the condition


 * $$\frac

= \frac $$

must be true, where $$\overline M_y = h_yM + hM_y$$ and $$\overline N_x = h_x + hN_x$$

Part A

In the Euler Integrating Factor Method, distributing $$ \displaystyle h(x,y) $$ in Eq. (8-3) gives:


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

$$\displaystyle \ \underbrace{h(x,y)M(x,y)}_{\overline{M}(x,y)}+\underbrace{h(x,y)N(x,y)}_{\overline{N}(x,y)}y'=0 $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 8-5)
 * }
 * }

The function $$\displaystyle h(x) $$ is used to make Eq. (8-3) exact, such that it satisfies the second condition of exactness, from Lecture 8, p. 8-4:


 * $$ \displaystyle \overline{M}_y(x,y)=\overline{N}_x(x,y) $$

Thus, from Eq. (8-5), we have:


 * $$ \displaystyle \not{h_y}M+hM_y= h_xN+hN_x $$
 * (Recall: $$ \displaystyle h_y=0 $$)
 * $$ \displaystyle h_xN+ h \left(N_x-M_y \right )=0 $$

Substitute $$ \displaystyle h_x $$ from Eq. (8-2):


 * $$ \displaystyle \frac{\partial}{\partial x} \left[e^{\int a_0\left( x \right)b\left( x \right)dx+k_1}\right] N+e^{\int a_0\left( x \right)b\left( x \right)dx+k_1} \left(N_x-M_y \right )=0 $$


 * $$ \displaystyle a_0(x) e^{\int a_0\left( x \right)b\left( x \right)dx+k_1} N+e^{\int a_0\left( x \right)b\left( x \right)dx+k_1} \left(N_x-M_y \right )=0 $$

Factor out $$ \displaystyle e^{\int a_0\left( x \right)b\left( x \right)dx+k_1} $$:


 * $$ \displaystyle e^{\int a_0\left( x \right)b\left( x \right)dx+k_1} \left[a_0(x)+N_x-M_y \right]=0 $$


 * $$ \displaystyle a_0(x)+N_x-M_y =0 $$


 * $$ \displaystyle $$

which we have rewritten


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

We have further substituted the term $$ \displaystyle h(x) $$ to represent our integrating factor, where King does not. By a simple substitution of terms, it can be shown that our results are consistent with King.


 * $$ \displaystyle h(x) = \exp^{\int^{x} P(t) dt} $$


 * $$\displaystyle b(x) = Q(x) $$


 * $$ \displaystyle a_0 (x) = P(x) $$

Our solution $$\displaystyle y(x) = \frac \int^{x} h(s) b(s) ds $$, becomes


 * $$ \displaystyle y(x) = e^{ - \int^{x} P(s) ds } \int^{x} h(s) b(s) ds $$

This is identical to the solution presented on page 512 in King [1]:


 * $$ \displaystyle y_p = \exp \left[ - \int^{x} P(t) dt \right] \int^{x} \underbrace{Q(s)}_{b(s)} \underbrace{\exp \left[ \int^{s} P(t) dt \right]}_{h(s)} ds$$

Part C

Our integrating factor is given by Eq. (8-2):


 * $$ \displaystyle h(x) = e^{\int^{x} a_0(s) ds} $$

To find the homogenous differential equation, start with the given equation:


 * $$ \displaystyle y' + a_0 y = 0 $$

Multiplying both sides by $$ \displaystyle h $$, yields


 * $$ \displaystyle h y' + a_0 h y = 0 $$

But the derivate of $$ \displaystyle h $$ with respect to $$ \displaystyle x $$ is:
 * $$ \displaystyle \frac{dh}{dx} = h' = \frac{d}{dx} e^{\int^{x} a_0(s) ds} = a_0 e^{\int^{x} a_0(s) ds } = a_0 h $$

Therefore we can write:


 * $$ \displaystyle h y' + h' y = 0 $$
 * $$ \displaystyle \frac{d}{dx} \left[ h(x) y(x) \right] = 0 $$
 * $$ \displaystyle h y' + h' y = 0 $$

By integrating both sides:


 * $$ \displaystyle \int^{x} (hy)' dx= \int^{x} 0 dx$$


 * $$ \displaystyle h(x) y_H(x) = A $$

where $$ \displaystyle A $$ is a constant of integration and $$ \displaystyle y_H(x) $$ is the homogenous solution of $$ \displaystyle y(x) $$.

Therefore,


 * $$ \displaystyle y_H(x) = \frac $$


 * $$ \displaystyle y_H(x) = {A}{e^{- \int^{x} a_0(s) ds}}$$

References:

68.17.105.125 05:52, 22 September 2010 (UTC)Egm6321.f10.team4.osentowski - Editor Part A, Author Part B and C 128.227.34.126 14:54, 22 September 2010 (UTC) - Author Part A, Editor Parts B and C

= Problem 9 - Making a N1-ODE exact =

From Meeting 12, p. 12-2.

Given
$$\displaystyle a(x) = \sin (x^3 )$$ $$\displaystyle b(x) = \cos (x) $$ $$\displaystyle c(y) = e^{2y} $$

Find
Part A

Find a N1-ODE that is either exact or can be made exact by using


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

$$
 * $$\displaystyle \bar b(x)c(y)y' + a(x)\bar c(y) = 0$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 9-1)
 * <p style="text-align:right;">$$\displaystyle (Eq. 9-1)
 * }
 * }

Part B

Find the first integral


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

$$
 * $$\displaystyle \phi (x,y) = k$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 9-2)
 * <p style="text-align:right;">$$\displaystyle (Eq. 9-2)
 * }
 * }

Solution
Part A

Lets first compute


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

$$
 * $$\displaystyle\bar b(x) = \int\limits_{}^x {b(s)ds} $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 9-5)
 * <p style="text-align:right;">$$\displaystyle (Eq. 9-5)
 * }
 * }

Which will yield the following


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

$$
 * $$\displaystyle\bar b(x) = \int\limits_{}^x {\cos (s)ds} = \sin (x)$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 9-6)
 * <p style="text-align:right;">$$\displaystyle (Eq. 9-6)
 * }
 * }

Doing the same for c


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

\bar c(y) = \int\limits_{}^y {c(s)ds} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 9-7)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

\bar c(y) = \int\limits_{}^y {e^{2s} ds} = \frac{1} {2}e^{2y} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 9-8)
 * }
 * }

Combining all into Eq. (9-1)


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

\underbrace {\sin (x)e^{2y} }_Ny' + \underbrace {\sin (x^3 )\frac{1} {2}e^{2y} }_M = 0$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 9-9)
 * }
 * }

One can see that Eq. (9-9) clearly satisfies the first condition of exactness. Lets now examine if the second condition shown in (Eq 4-2) is satisfied.


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

\frac = \frac{\partial } \left( {\sin (x^3 )\frac{1} {2}e^{2y} } \right) = \sin (x^3 )e^{2y} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 9-10)
 * }
 * }


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

\frac = \frac{\partial } \left( {\sin (x)\frac{1} {2}e^{2y} } \right) = \cos (x)e^{2y} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 9-11)
 * }
 * }

Comparing the above two equations it is clear that Eq. (9-9) will have to be made exact by implementing Euler's Integrating Factor Method. However, before doing so, let's assume $$e^{2y} \ne 0$$ and divide it through to get


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

{2}}_M = 0$$ $$
 * $$\displaystyle\underbrace {\sin (x)}_Ny' + \underbrace {\sin (x^3 )\frac{1}
 * $$\displaystyle\underbrace {\sin (x)}_Ny' + \underbrace {\sin (x^3 )\frac{1}
 * <p style="text-align:right;">$$\displaystyle (Eq. 9-12)
 * }
 * }

Multiply everything by an integration factor h.


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

$$
 * $$\displaystyle\underbrace {hN}_{\bar N}y' + \underbrace {hM}_{\bar M} = 0$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 9-13)
 * <p style="text-align:right;">$$\displaystyle (Eq. 9-13)
 * }
 * }

The above (Eq. (9-13)) still satisfies the first condition of exactness, but we are going to force the second one:


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

= \frac $$ $$
 * $$\displaystyle \frac
 * $$\displaystyle \frac
 * <p style="text-align:right;">$$\displaystyle (Eq. 9-14)
 * }
 * }

and expanding:


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

h_x N + hN_x = h_y M + hM_y $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 9-15)
 * }
 * }

By inspection of Eq. (9-12), we see that both $$\displaystyle N $$ and $$\displaystyle M $$ are only functions of $$\displaystyle x $$. Therefore, we will assume that $$\displaystyle h $$ is also a function of $$\displaystyle x $$ only to get the following:


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

\frac {h} = \frac{1} {N}\left( {M_y - N_x } \right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 9-16)
 * }
 * }

Integrating Eq. (9-16)


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

\ln \left| {h(x)} \right| = \int {\frac{1} (0 - \cos (x))dx = - \ln \left| {\sin (x)} \right|} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 9-17)
 * }
 * }


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

h(x) = e^{ - \ln \left| {\sin (x)} \right|} = \frac{1} = \frac{1} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 9-18)
 * }
 * }

The exact N1-ODE will be


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

y' + \frac = 0$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 9-19)
 * }
 * }

Part B

We know that


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

= \underbrace {\frac }_M + \underbrace {\frac }_N\underbrace {\frac }_{y'} $$ $$
 * $$\displaystyle F(x,y,y') = \frac
 * $$\displaystyle F(x,y,y') = \frac
 * <p style="text-align:right;">$$\displaystyle (Eq. 9-20)
 * }
 * }

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

= M(x,y) $$ $$
 * $$\displaystyle \frac
 * $$\displaystyle \frac
 * <p style="text-align:right;">$$\displaystyle (Eq. 9-21)
 * }
 * }

and


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

= N(x,y) $$ $$
 * $$\displaystyle \frac
 * $$\displaystyle \frac
 * <p style="text-align:right;">$$\displaystyle (Eq. 9-22)
 * }
 * }

Integrating Eq. (9-21) with respect to $$\displaystyle x $$ and Eq. (9-22) with respect to $$\displaystyle y $$ we will get the following:


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

$$ $$
 * $$\displaystyle \int {\partial \phi } = \int {1dy = y + c_1 }
 * $$\displaystyle \int {\partial \phi } = \int {1dy = y + c_1 }
 * <p style="text-align:right;">$$\displaystyle (Eq. 9-23)
 * }
 * }


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

\int {\partial \phi } = \int {\frac dx = \frac{1} {2}} \int {\frac dx + c_2 } $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 9-24)
 * }
 * }

Combining all together:


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

$$\displaystyle \phi (x,y) = \frac{1} {2}\int {\frac dx + c_2 } + y + c_1  = \frac{1} {2}\int {\frac dx + } y + c$$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 9-25)
 * }
 * }

Egm6321.f10.team4.petralanda.n 03:05, 20 September 2010 (UTC) - Author