User:Egm6321.f09.team3.diggs/HW3

= Problem 1 =

Given
$$ x^{m+ \frac{1}{2}}y^ny'' + 2x^{1+m}y^ny' + 3x^my^{n+1} $$

Find
m & n such that the ODE is exact.

Solution
$$f(x,y,p) = x^{m+ \frac{1}{2}}y^n$$

$$g(x,y,p) = 2x^{1+m}y^ny' + 3x^my^{n+1} $$

which satisfies the first condition of exactness.

Condition 2 is met by satifaction of the following equations:

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

ii. $$ f_{xp} + pf_{yp} + 2f_y = g_{pp} $$

The partial derivatives are given below:

$$ f_x = (m+ \frac{1}{2})x^{m- \frac{1}{2}}y^n$$

$$ f_{xx} = (m- \frac{1}{2}) (m+ \frac{1}{2})x^{m- \frac{3}{2}}y^n$$

$$ f_{xy} = n(m+ \frac{1}{2})x^{m- \frac{1}{2}}y^{n-1}$$

$$ f_y = nx^{m+ \frac{1}{2}}y^{n-1} $$

$$ f_{yy} = n(n-1)x^{m+ \frac{1}{2}}y^{n-2} $$

$$ f_p = f_{xp} = f_{yp} = f_{pp} = 0 $$

$$ g_x = 2(m+1)x^my^np + 3mx^{m-1}y^{n+1} $$

$$ g_{xx} = 2m(m+1)x^{m-1}y^np + 3m(m-1)x^{m-2}y^{n+1} $$

$$ g_{xy} = 2(m+1)nx^my^{n-1}p + 3m(n+1)x^{m-1}y^n $$

$$ g_y = 2nx^{m+1}y^{n-1}p + 3(n+1)x^my^n $$

$$ g_{yy} = 2n(n-1)2nx^{m+1}y^{n-2}p + 3n(n+1)x^my^{n-1} $$

$$ g_p = 2x^{m+1}y^n $$ $$ g_{px} = 2(m+1)x^my^n $$

$$ g_{py} = 2nx^{m+1}y^{n-1} $$

$$ g_{pp} = 0 $$

Substitute into i. to get:

$$ (m- \frac{1}{2}) (m+ \frac{1}{2})x^{m- \frac{3}{2}}y^n + 2p \cdot n(m+ \frac{1}{2})x^{m- \frac{1}{2}}y^{n-1} + p^2 \cdot n(n-1)x^{m+ \frac{1}{2}}y^{n-2} = 2(m+1)x^my^n + p \cdot 2nx^{m+1}y^{n-1} - 2nx^{m+1}y^{n-1}p + 3(n+1)x^my^n $$

Substitute into ii. to get:

$$ 0 + p \cdot 0 + 2 \cdot nx^{m+ \frac{1}{2}}y^{n-1} = 0 $$

Equation ii. suggests that n must be zero in order for the equation to be satisfied.

$$ n = 0 $$

Simplify equation i to get:

$$ (m- \frac{1}{2}) (m+ \frac{1}{2})x^{m- \frac{3}{2}} + 0 + 0 = 2(m+1)x^m + 0 - 0 + 3x^m $$

Note the powers of x need to be the same to simplify, which suggests $$ m = +/- \frac{1}{2} $$.

$$ m = \frac{1}{2} $$ satisfies equation i.

Therefore,

$$ n = 0 $$

and

$$ m = \frac{1}{2} $$

The final ODE is

$$ xy'' + 2x^{\frac{3}{2}}y' + 3x^{\frac{1}{2}}my $$

or,

$$ y'' + 2x^{\frac{1}{2}}y' + 3x^{\frac{-1}{2}}my $$

= Problem 2 =

Given
$$ xp - (2x^{\frac {3}{2}} - 1)y + k_1 = k_2 $$

Find
y(x)

Solution
The ODE is

$$ y' - \frac{2x^{\frac{3}{2}}-1}{x}y = k_2 - k_1 $$

where $$ a_0(x) = - \frac{2x^{\frac{3}{2}}-1}{x} $$ and $$ b(x) = k_2 - k_1 $$

This is a linear, first order ODE with varying coefficients.

In standard form, $$ N = 1 $$ and $$ M = a_0 \cdot y $$.

This makes $$ h(x) = exp( \int a_0(s) ds = exp( \int (-2 \sqrt{s} + s^{-1}) ds) = x \cdot exp(- \frac{4}{3} x^{\frac {3}{2}}) $$

Multiply the ODE by the integrating factor to get:

$$ (hy)' = hb $$

Integrate to get:

$$ y(x) = \frac{1}{h(x)} \int h(s) b(s) ds $$

$$ y(x) = \frac{1}{x \cdot exp(- \frac{4}{3} x^{\frac {3}{2}})} \int s \cdot exp(- \frac{4}{3} s^{\frac {3}{2}}) \cdot (k_2 - k_1) ds $$

NEED TO SOLVE THE INTEGRAL!!

= Problem 3 =

Given
A class of exact linear second order ODEs with varying coefficients

Find
The mathematical structure of $$\phi$$

Solution
Since $$ F = \frac {d \phi}{dx} = \phi_x(x,y,p) + \phi_y(x,y,p) \cdot y' + \phi_p(x,y,p) \cdot y' $$

and $$ F = P(x) y'' + Q(x) y' + R(x) y $$

This suggests

i. $$ P(x) = \phi_p $$ ii. $$ Q(x) = \phi_y $$

iii. $$ R(x)y = \phi_x $$

From i.,

$$ \phi = \int P(x) dp = P(x) \cdot p + C(x,y) $$

From ii.,

$$ \phi = \int Q(x) dy = Q(x) \cdot y + C(x,p) $$

This suggests $$ \phi $$ has a structure of:

$$ \phi = P(x) \cdot p + Q(x) \cdot y + C(x,y) $$

Take the parital derivative with respect to x,

$$ \frac{\partial{\phi}}{dx} = \frac{\partial{P}}{dx} \cdot p + \frac{\partial{Q}}{dx} \cdot y + \frac{\partial{C}}{dx} $$

and compare to iii, which suggests C is not a function of x!

The final form is:

$$ \phi = P(x) \cdot p + T(x) \cdot y + k $$

where P(x) and T(x) are functions, and k is a constant.

= Problem 4 =

Given
The conditions for exactness:

i. $$ F = \frac{d\phi}{dx} = \phi_x + \phi_y \cdot y' + ... + \phi_{y^{n-1}} \cdot y^n $$

ii. $$ f_i = \partial{F}{y^{(i)}} $$ such that $$ f_0 - \frac{df_1}{dx} + \frac{d^2f_2}{dx^2} - ... + (-1)^n \frac{d^nf_n}{dx^n} = 0 $$

Find
the n= 1 case

Solution
$$ f_1 = \frac{\partial{F}}{dy'} = \frac{\partial{}}{dy'}(\phi_x + \phi_y y') = \phi_y $$

Therefore,

$$ f_0 - \frac{df_1}{dx} = \frac{\partial {F}}{dy} - \frac{d}{dy}(\phi_y) = \frac{\partial{(\phi_x + \phi_y y')}}{dy'} - \frac{d}{dx}(\phi_y) $$

$$ \frac{\partial{}}{dy'}(\phi_x + \phi_y y') - \frac{d}{dx}(\phi_y) = \phi_{xy} + \phi_{yy}y' - \phi_yx - \phi_{yy}y' $$

$$ \frac{\partial{}}{dy'}(\phi_x + \phi_y y') - \frac{d}{dx}(\phi_y) = \phi_{xy}  - \phi_yx $$

Therefore,

$$ f_0 - \frac{df_1}{dx} = 0 $$ is equivalent to $$ \phi_{xy}  = \phi_yx $$, which is the condition of exactness for the linear first order ODE!

= Problem 5 =

Given
The conditions for exactness:

i. $$ F = \frac{d\phi}{dx} = \phi_x + \phi_y \cdot y' + ... + \phi_{y^{n-1}} \cdot y^n $$

ii. $$ f_i = \partial{F}{y^{(i)}} $$ such that $$ f_0 - \frac{df_1}{dx} + \frac{d^2f_2}{dx^2} - ... + (-1)^n \frac{d^nf_n}{dx^n} = 0 $$

Find
the n=2 case

Solution
$$ f_0 - \frac{df_1}{dx} + \frac{d^2f_2}{dx^2} $$

Integrate with respect to x to get:

$$ \int f_0 dx - f_1 + \frac{df_2}{dx} $$

where $$ f_0 = \frac{\partial{F}}{dy} = \phi_{xy} + \phi_{yy} y' + \phi_{y'y} y'' $$

Therefore, $$ \int f_0 dx = \phi_y $$

Now, take the derivative with respect to x to get the original equation,

$$ \frac{d}{dx}(\phi_y) = \phi_{xy} + \phi_{yy} y' + \phi_{yy'} y'' $$

which is, of course, $$ f_0 $$!

The final equations is validated:

$$ f_0 - \frac{df_1}{dx} + \frac{d^2f_2}{dx^2} $$

= Problem 6 =

Given
The Legendre linear, second order ODE with varying coefficients: $$ F = (1-x^2)y'' - 2xy' + n(n+1)y = 0 $$

Find
Verify the exactness of the equation using:

a. equations 4 and 5 from 10-2

b. equation 5 from 14-1

Solution
a. Using the following 2 equations:

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

ii. $$ f_{xp} + pf_{yp} + 2 f_y = g_{pp} $$

where F = f(x,y,p)y'' + g(x,y,p)

$$f(x,y,p) = 1-x^2$$

$$g(x,y,p) = -2xp + n(n-1) y$$

The partial derivatives are:

$$f_x = -2x$$

$$f_{xx} = -2$$

$$f_{xy} = f_y = f_{yy} = f_{xp} = f_{yp} = f_p = f_{pp} = 0$$

$$g_x = -2p$$

$$g_{xp} = -2$$

$$g_y = n(n+1)$$

$$g_p = -2x$$

$$ g_{xx} = g_{xy} = g_{yy} = g_{yp} = g_{pp} = 0 $$

Substitute to get:

i. $$ -2 + 2p \cdot 0 + p^2 \cdot 0 = -2 + p \cdot 0 - n(n+1) $$

ii. $$ 0 + p \cdot 0 + 2 \cdot 0 = 0 $$

The ODE is not exact because the equation i is not satisfied.

b. Using the following equation:

$$ f_0 - \frac{df_1}{dx} + \frac{d^2f_2}{dx^2} = 0 $$

where

$$ f_0 = \frac{\partial{F}}{dy} = n(n+1) $$

$$ f_1 = \frac{\partial{F}}{dy'} = -2x $$

$$ f_2 = \frac{\partial{F}}{dy''} = 1-x^2 $$

Combining,

$$n(n+1) - (-2) + (-2) = 0$$

Which is exact for the solution of $$ n = 0 $$ (trivial) or $$ n=-1 $$

However, the ODE can be made exact using the integrating factors method:

$$h(x,y) = x^my^n$$

Multiply by the ODEto get:

$$ x^my^n [ (1-x^2)y'' - 2xy' + n(n+1) y = 0 ] $$

which makes $$ f(x,y,p) = x^my^n - x^{m+2}y^n $$ and $$ g(x,y,p) = - 2x^{m+1}y^np + \alpha(\alpha+1)x^m y^{n+1} $$

The partial derivatives are:

$$f_x = mx^{m-1}y^n - (m+2)x^{m+1}y^n$$

$$f_{xx} = m(m-1)x^{m-2}y^n - (m+2)(m+1)x^{m}y^n$$

$$f_{xy} = mnx^{m-1}y^{n-1} - (m+2)nx^{m+1}y^{n-1}$$

$$f_y = nx^my^{n-1} - nx^{m+2}y^{n-1} $$

$$ f_{yy} = n(n-1)x^my^{n-2} - n(n-1)x^{m+2}y^{n-2} $$

$$ f_{xp} = f_{yp} = f_p = f_{pp} = 0 $$

$$ g_x = -2(m+1)x^my^np + m\alpha (\alpha + 1)x^{m-1}y^{n+1} $$

$$ g_{xx} = -2m(m+1)x^{m-1}y^np + m(m-1)\alpha (\alpha + 1)x^{m-2}y^{n+1} $$

$$ g_{xy} = -2(m+1)nx^my^{n-1}p + m\alpha (\alpha + 1)(n+1)x^{m-1}y^{n} $$

$$ g_y = - 2nx^{m+1}y^{n-1}p + (n+1)\alpha(\alpha+1)x^m y^{n} $$

$$ g_{yy} = - 2n(n-1)x^{m+1}y^{n-2}p + n(n+1)\alpha(\alpha+1)x^m y^{n-1} $$

$$ g_p = -2x^{m+1}y^n $$

$$ g_{yp} = -2nx^{m+1}y^{n-1} $$

$$ g_{xp} = -2(m+1)x^my^n $$

$$ g_{pp} = 0 $$

Substituting into equation i:

$$ m(m-1)x^{m-2}y^n - (m+2)(m+1)x^{m}y^n + 2p \cdot (mnx^{m-1}y^{n-1} - (m+2)nx^{m+1}y^{n-1}) + p^2 \cdot (n(n-1)x^my^{n-2} - n(n-1)x^{m+2}y^{n-2}) = -2(m+1)x^my^n + p \cdot (-2nx^{m+1}y^{n-1}) - (- 2nx^{m+1}y^{n-1}p) + (n+1)\alpha(\alpha+1)x^m y^{n}) $$

and into equation ii:

$$ 0 + p \cdot 0 + 2 \cdot (nx^my^{n-1} - nx^{m+2}y^{n-1}) = 0 $$

Equation ii can only be solved by $$ n = 0 $$

which leaves equation i as:

$$ m(m-1)x^{m-2} - (m+2)(m+1)x^{m} + 2p \cdot (0 + 0) + p^2 \cdot (0 - 0) = -2(m+1)x^m + p \cdot (0) - (-0) + \alpha(\alpha+1)x^m) $$

Which may be satisfied for $$ m = 1 $$

(Incidentally, the solution requires $$ \alpha = 1 $$ to be true.)

= Problem 7 =

Given
The definitions of linearity:

a. $$ L(\alpha u + \beta v) = \alpha L(u) + \beta L(v)$$

b. i. $$ L(u + v) = L(u) + L(v) $$ and ii. $$ L( \lambda u ) = \lambda L(u) $$

Find
Show the two definitions are true.

Solution
Apply b. ii to get:

$$ L( \alpha u ) = \alpha L(u)$$

and

$$ L( \beta v ) = \beta L(v) $$

Combine using b. i to get:

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

which matches the definition of linearity given in a.

= Problem 8 =

CAN SOMEONE UPLOAD A SCANNED DRAWING HERE?

= Problem 9 =

Given
x^2 y'' - 2xy' + 2y = 0

Find
y_{xxx} and y_{xxxx}

Solution
Recall that $$ \frac{d}{dt} y = \frac{dy}{dx} \cdot \frac{dx}{dt} = \frac{dy}{dx} \cdot e^t $$

Which yields $$ y_t = y_x \cdot e^t $$

Rearrange to get: $$ y_x = y_t \cdot e^{-t} $$

Take the third derivative:

$$ \frac{d^3}{dx^3}y = \frac{d}{dx}(\frac{d}{dx}(\frac{d}{dx}(y))) = \frac{dt}{dx}\frac{d}{dt}(\frac{dt}{dx}\frac{d}{dt}(\frac{dt}{dx}\frac{d}{dt}(y))) $$

$$ \frac{d^3}{dx^3}y = e^{-t}\frac{d}{dt}(e^{-t}\frac{d}{dt}(e^{-t}\frac{d}{dt}(y))) $$

$$ \frac{d^3}{dx^3}y = e^{-t}\frac{d}{dt}(e^{-t}\frac{d}{dt}(e^{-t} y_t)) $$

$$ \frac{d^3}{dx^3}y = e^{-t}\frac{d}{dt}(e^{-t}(-e^{-t} y_t + e^{-t}y_{tt})) $$

$$ \frac{d^3}{dx^3}y = e^{-t}\frac{d}{dt}(-e^{-2t} y_t + e^{-2t}y_{tt}) $$

$$ \frac{d^3}{dx^3}y = e^{-t}(2e^{-2t} y_t - e^{-2t}y_{tt} - 2e^{-2t}y_{tt} + e^{-2t}y_{ttt}) $$

$$ \frac{d^3}{dx^3}y = 2e^{-3t} y_t - 3e^{-3t}y_{tt} + e^{-3t}y_{ttt} $$

$$ y_{xxx} = e^{-3t} (2y_t - 3y_{tt} + y_{ttt}) $$

To find $$ y_{xxxx} $$ take an additional derivative:

$$ y_{xxxx} = \frac{dt}{dx}\frac{d}{dt}(2e^{-3t} y_t - 3e^{-3t}y_{tt} + e^{-3t}y_{ttt}) $$

$$ y_{xxxx} = e^{-t} (-3 e^{-3t}y_{ttt} + e^{-3t}y_{tttt} + 9 e^{-3t}y_{tt} - 3 e^{-3t}y_{ttt} - 6 e^{-3t}y_t + 2e^{-3t}y_{tt} $$

$$ y_{xxxx} = e^{-4t}(y_{tttt} - 6y_{ttt} + 11y_{tt} - 6y_t) $$

= Problem 10 =

Given
x^2y'' - 2xy' + 2y = 0

Find
The solution using the method of trial solution $$ y = e^{rx} $$

Solution
= Problem 11 =

Given
$$ u_1(x) z' + [a_1(x)u_1(x) + 2u_1'(x)]z = 0 $$

Find
Solution using the integrating factor method.

Solution
$$ z' + [a_1(x) + 2 \frac {u_1'(x)}{u_1(x)}]z = 0 $$

$$ a_0 = a_1 + 2 \frac {u_1'}{u_1}] $$ and $$ b = 0 $$

The integrating factor h(x) is:

$$ h(x) = exp( \int a_0(s) ds = exp( \int a1(s) + 2\frac {u_1'(s)}{u_1(s)} ds) = exp( \int a1(s) ds + 2 ln(x)) $$

$$ (hz)' = 0 $$

Integrate to solve:

$$ hz = constant $$

Rearrange to solve for z:

$$ z = \frac {constant}{(u_1(x))^2} \cdot exp[- \int a_1(s) ds] $$

= Problem 12 =

Given
The reduction of order method 2 using $$ y(x) = U(x)u_1(x) $$

Find
Alternate definitions of y(x) that create reduction of order method 2

a. $$ y(x) = U(x) +/- u_1(x) $$

b. $$ y(x) = U(x)/u_1(x) $$

c. $$ y(x) = u_1(x)/U(x) $$

Solution
a.

Take the first and second derivatives of y(x) to get:

$$y'(x) = U' +/- u_1'$$

$$y(x) = U +/- u_1''$$

The desired form is $$ y'' + a_1y' + a_0y = 0 $$, multiply the above equations appropriately to get:

$$ a_0y + a_1y' + y = a_0U +/- a_0u_1 + a_0U' +/- a_1u_a' + U +/- u_1'' $$

Which is $$ 0 = a_0U +/- a_0u_1 + a_0U' +/- a_1u_a' + U +/- u_1 $$

The assumed form of y(x) here is not valid because the equation above is not missing U. The method cannot be developed further.

b.

Take the first and second derivatives of y(x) to get:

$$y'(x) = U'/u_1 + U/u_1'$$

$$y(x) = U/u_1 + 2U'/u_1' + U/u_1''$$

The desired form is $$ y'' + a_1y' + a_0y = 0 $$, multiply the above equations appropriately to get:

$$ a_0y + a_1y' + y = a_0 U/u_1 + a_1 U'/u_1 + a_1 U/u_1' + U/u_1 + 2U'/u_1' + U/u_1'' $$

Which is $$ 0 = a_0 U/u_1 + a_1 U'/u_1 + a_1 U/u_1' + U/u_1 + 2U'/u_1' + U/u_1 $$

The assumed form of y(x) here is not valid because the equation above is not missing U. The method cannot be developed further.

c.

Take the first and second derivatives of y(x) to get:

$$y'(x) = u_1'/U + u_1/U'$$

$$y(x) = u_1/U + 2u_1'/U' + u_1/U''$$

The desired form is $$ y'' + a_1y' + a_0y = 0 $$, multiply the above equations appropriately to get:

$$ a_0y + a_1y' + y = a_0 u_1/U + a_1 u_1'/U + a_1 u_1/U'+ u_1/U + 2u_1'/U' + u_1/U''$$

Which is $$ 0 = a_0 u_1/U + a_1 u_1'/U + a_1 u_1/U'+ u_1/U + 2u_1'/U' + u_1/U $$

The assumed form of y(x) here is not valid because the equation above is not missing U. The method cannot be developed further.