User:Egm4313.s12.team19.R2

R2.1
Contributed by Vladimir Horwitz.

Problem Statement:
a) Find a nonhomogeneous L2ODECC for the roots and initial conditions listed here
 * {| style = "width: 100%"

!  $$ \lambda_1 = - 2, \ \lambda_2 = 5 $$ !!  (Eq. 1, a and b) The L2ODECC should be of standard form and be written for the general excitation r(x).
 *  $$ \ y (0) = 1, \ y' (0) = 0 $$
 *  (Eq. 2, a and b)
 * }
 * }

b) Plot the solution for $$ \ r (x) = 0 $$.

c) Write 3 L2ODECCs that are not standard and not homogeneous that will accept (eq. 1, a and b) and (eq. 2, a and b).

Solution:
a) The characteristic equation for the given roots and initial conditions is of the form $$ \ (\lambda - \lambda_1) * (\lambda - \lambda_2) = 0 $$.

Therefore,


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!  $$ \ y'' - 3 y' - 10 y = r (x) $$  (Eq. 3)
 * }

The solutions should be of the form $$ \ y (x) = y_p (x) + y_h (x) $$.

Where according to (eq. 1, a and b) and (eq. 2, a and b)


 * {| style = "width: 100%"

!  $$ \ y_h (0) = c_1 + c_2 $$ !!  (Eq. 4)
 *  $$ \ y'_h (0) = c_1 \lambda_1 + c_2 \lambda_2 $$
 *  (Eq. 5)
 * and
 *  $$ \ c_1 = {5 \over 7} = .71429 $$
 *  (Eq. 6)
 *  $$ \ c_2 = {2 \over 7} = .28571 $$
 *  (Eq. 7)
 * Thus,
 *  $$ \ y(x) = \overbrace {.71429 e^{(-2 x)} + .28571 e^{(5 x)}}^{y_h (x)} + y_p (x) $$
 *  (Eq. 8)
 * }
 * Thus,
 * <p style = "text-align:left"> $$ \ y(x) = \overbrace {.71429 e^{(-2 x)} + .28571 e^{(5 x)}}^{y_h (x)} + y_p (x) $$
 * <p style = "text-align:right"> (Eq. 8)
 * }
 * <p style = "text-align:right"> (Eq. 8)
 * }

b) We can now plot the homogeneous form of (eq. 8) $$ \underbrace {y_h (x) = .71429 e^{(-2 x)} + .28571 e^{(5 x)}}_{\downarrow} $$.

c) Three more equations can be attained for the given (eq. 1, a and b) and (eq. 2, a and b), because of the linearity.


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! <p style = "text-align:left"> $$ \ 3 y'' - 9 y' - 30 y = r (x) $$ <p style = "text-align:right"> (Eq. 9)
 * }


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! <p style = "text-align:left"> $$ \ 10 y'' - 30 y' - 100 y = r (x) $$ <p style = "text-align:right"> (Eq. 10)
 * }


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! <p style = "text-align:left"> $$ \ 5 y'' - 15 y' - 50 y = r (x) $$ <p style = "text-align:right"> (Eq. 11)
 * }

R2.2
Contributed by Gabriel Arab.

Problem Statement:
Find and plot the solution for the L2-ODE-CC,


 * {| style = "width: 100%"

! <p style = "text-align:left"> $$ \ y'' - 10 y' + 25 y = r (x) $$ !! <p style = "text-align:right"> (Eq. 1)
 * Initial conditions.
 * <p style = "text-align:left"> $$ \ y (0) = 1, \ y' (0) = 0 $$
 * <p style = "text-align:right"> (Eq. 2)
 * No excitation.
 * <p style = "text-align:left"> $$ \ r (x) = 0 $$
 * <p style = "text-align:right"> (Eq. 3)
 * }
 * <p style = "text-align:left"> $$ \ r (x) = 0 $$
 * <p style = "text-align:right"> (Eq. 3)
 * }
 * }

Solution:
Converting original ODE into λ notation and factoring,


 * {| style = "width: 100%"

!<p style = "text-align:left"> $$ \ \lambda^2 - 10 \lambda + 25 = r (x) $$ !! <p style = "text-align:right"> (Eq. 4)
 * <p style = "text-align:left"> $$ \ (\lambda - 5)^2 = r (x) = 0 $$
 * <p style = "text-align:right"> (Eq. 5)
 * We find that
 * <p style = "text-align:left"> $$ \ \lambda = 5 $$ is a double real root.
 * <p style = "text-align:right"> (Eq. 6)
 * }
 * <p style = "text-align:left"> $$ \ \lambda = 5 $$ is a double real root.
 * <p style = "text-align:right"> (Eq. 6)
 * }

Implementing the solution equation for real double roots and deriving,


 * {| style = "width: 100%"

! <p style = "text-align:left"> $$ \ y_h (x) = c_1 e^{5 x} + c_2 x e^{5 x} $$ !! <p style = "text-align:right"> (Eq. 7)
 * <p style = "text-align:left"> $$ \ y_h' (x) = 5 c_1 e^{5 x} + 5 c_2 x e^{5 x} + c_2 e^{5 x} $$
 * <p style = "text-align:right"> (Eq. 8)
 * }
 * }

Plugging in initial conditions,


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! <p style = "text-align:left"> $$ \ y_h (0) = c_1 e^{5 (0)} + c_2 (0) e^{5 (0)} = 1 $$ !! <p style = "text-align:right"> (Eq. 9)
 * <p style = "text-align:left"> $$ \ 1 = c_1 $$
 * <p style = "text-align:right"> (Eq. 10)
 * and
 * <p style = "text-align:left"> $$ \ y_h' (0) = 5 c_1 e^{5 (0)} + 5 c_2 (0) e^{5 (0)} + c_2 e^{5 (0)} = 0 $$
 * <p style = "text-align:right"> (Eq. 11)
 * <p style = "text-align:left"> $$ \ 5 c_1 + c_2 = 0 $$
 * <p style = "text-align:right"> (Eq. 12)
 * Thus,
 * <p style = "text-align:left"> $$ \ c_2 = -5 $$
 * <p style = "text-align:right"> (Eq. 13)
 * }
 * Thus,
 * <p style = "text-align:left"> $$ \ c_2 = -5 $$
 * <p style = "text-align:right"> (Eq. 13)
 * }
 * <p style = "text-align:right"> (Eq. 13)
 * }

Writing final solution equation and plotting on WolframAlpha engine,


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! <p style = "text-align:left"> $$ \ y_h (x) = e^{5 x} - 5 x e^{5 x} $$ <p style = "text-align:right"> (Eq. 14)
 * }

R2.3
Contributed by Adam Burton.

Given:

Problem Statement:
From Kreyszig 2011 page 59, problems 3 and 4.

Find a general solution. Check your answers by substitution.

Solution:
3) :{| style = "width: 100%" ! <p style = "text-align:left"> $$ \ y'' + 6y' + 8.96y = 0  $$ !! <p style = "text-align:right"> (Eq. 1)
 * }

Substituting

! <p style = "text-align:left"> $$ \ h = d/dx  $$ !! <p style = "text-align:right"> (Eq. 2)
 * }


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! <p style = "text-align:left"> $$ \ h^2 + 6h + 8.96 = 0 $$ !! <p style = "text-align:right"> (Eq. 3)
 * }

To solve this, we will use the quadratic formula:


 * {| style = "width: 100%"

! <p style = "text-align:left"> $$ \ h = \frac{-b \plusmn\! \sqrt{b^2-4ac}}{2a} $$ !! <p style = "text-align:right"> (Eq. 4)
 * }


 * {| style = "width: 100%"

! <p style = "text-align:left"> $$ \ h = \frac{-6 \plusmn\! \sqrt{6^2-4(1)(8.96)}}{2(1)} $$ !! <p style = "text-align:right"> (Eq. 4a)
 * }

Through calculation, we see that:


 * {| style = "width: 100%"

! <p style = "text-align:left"> $$ \ h_{1,2} = -2.8, -3.2$$ !! <p style = "text-align:right"> (Eq. 5)
 * }

Therefore, the general solution is:


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! <p style = "text-align:left"> $$ \ y = c_1e^{-2.8x} + c_2e^{-3.2x} $$ <p style = "text-align:right"> (Eq. 6)
 * }

Check:


 * {| style = "width: 100%"

! <p style = "text-align:left"> $$ \ y = c_1e^{-2.8x} + c_2e^{-3.2x} $$ !! <p style = "text-align:right"> (Eq. 6)
 * }


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! <p style = "text-align:left"> $$ \ y' = -2.8c_1e^{-2.8x} - 3.2c_2e^{-3.2x} $$ !! <p style = "text-align:right"> (Eq. 7)
 * }


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! <p style = "text-align:left"> $$ \ y'' = 7.84c_1e^{-2.8x} + 10.24c_2e^{-3.2x} $$ !! <p style = "text-align:right"> (Eq. 8)
 * }

Substituting these 3 equations in to equation 1 we get:


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! <p style = "text-align:left"> $$ \ 7.84c_1e^{-2.8x} + 10.24c_2e^{-3.2x} + 6(-2.8c_1e^{-2.8x} - 3.2c_2e^{-3.2x}) + 8.96(c_1e^{-2.8x} + c_2e^{-3.2x}) = 0 $$ !! <p style = "text-align:right"> (Eq. 9)
 * }


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! <p style = "text-align:left"> $$ \ 0 = 0 $$ !! <p style = "text-align:right"> (Eq. 10)
 * }

Through substitution, we can see that the solution above is correct!

4)
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! <p style = "text-align:left"> $$ \ y'' + 4y' + (\pi^2 + 4)y = 0 $$ !! <p style = "text-align:right"> (Eq. 1)
 * }

Substituting

! <p style = "text-align:left"> $$ \ h = d/dx $$ !! <p style = "text-align:right"> (Eq. 2)
 * }


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! <p style = "text-align:left"> $$ \ h^2 + 4h + (\pi^2+4) = 0 $$ !! <p style = "text-align:right"> (Eq. 3)
 * }

To solve this, we will use the quadratic formula:


 * {| style = "width: 100%"

! <p style = "text-align:left"> $$ \ h = \frac{-b \plusmn\! \sqrt{b^2-4ac}}{2a} $$ !! <p style = "text-align:right"> (Eq. 4)
 * }


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! <p style = "text-align:left"> $$ \ h = \frac{-4 \plusmn\! \sqrt{4^2-4(1)(\pi^2+4)}}{2(1)} $$ !! <p style = "text-align:right"> (Eq. 4a)
 * }


 * {| style = "width: 100%"

! <p style = "text-align:left"> $$ \ h = \frac{-4 \plusmn\! \sqrt{2\pi}}{2(1)} $$ !! <p style = "text-align:right"> (Eq. 4b)
 * }

Through calculation, we see that:


 * {| style = "width: 100%"

! <p style = "text-align:left"> $$ \ h_{1,2} = -2 \plusmn\ i\pi$$ !! <p style = "text-align:right"> (Eq. 5)
 * }

Therefore, the general solution is:


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! <p style = "text-align:left"> $$ \ y = e^{-2x}(c_1\cos\pi x + c_2\sin\pi x) $$ <p style = "text-align:right"> (Eq. 6)
 * }

Check:


 * {| style = "width: 100%"

! <p style = "text-align:left"> $$ \  y = e^{-2x}(c_1\cos\pi x + c_2\sin\pi x)  $$ !! <p style = "text-align:right"> (Eq. 6)
 * }


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! <p style = "text-align:left"> $$ \  y' = e^{-2x}(-\pi c_1\sin\pi x + \pi c_2\cos\pi x) - 2e^{-2x}(c_1\cos\pi x + c_2\sin\pi x) $$ !! <p style = "text-align:right"> (Eq. 7)
 * }


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! <p style = "text-align:left"> $$ \  y'' = 9e^{-2x}(c_1\cos\pi x + c_2\sin\pi x) + e^{-2x}(-\pi c_1\cos\pi x + \pi c_2\sin\pi x) - 6e^{-2x}(-\pi c_1\sin\pi x + \pi c_2\cos\pi x) $$ !! <p style = "text-align:right"> (Eq. 8)
 * }

Substituting these 3 equations in to equation 1 we get:


 * {| style = "width: 100%"

! <p style = "text-align:left"> $$ \  9e^{-2x}(c_1\cos\pi x + c_2\sin\pi x) + e^{-2x}(-\pi c_1\cos\pi x + \pi c_2\sin\pi x) - 6e^{-2x}(-\pi c_1\sin\pi x + \pi c_2\cos\pi x) \ + \ \downarrow $$
 * <p style = "text-align:left"> $$ \uparrow + \ 4(e^{-2x}(-\pi c_1\sin\pi x + \pi c_2\cos\pi x) - 2e^{-2x}(c_1\cos\pi x + c_2\sin\pi x)) + (\pi^2 + 4)(e^{-2x}(c_1\cos\pi x + c_2\sin\pi x)) = 0 $$
 * <p style = "text-align:right"> (Eq. 9)
 * }
 * }


 * {| style = "width: 100%"

! <p style = "text-align:left"> $$ \  0 = 0 $$ !! <p style = "text-align:right"> (Eq. 10)
 * }

Through substitution, we can see that the solution above is correct!

R2.4
Contributed by Joshua Campbell.

Problem 5

Given:


 * {| style = "width: 100%"

! <p style = "text-align:left"> $$ \ y'' + 2\pi y' + \pi^{2} y =0 $$ !! <p style = "text-align:right"> (Eq. 1)
 * }

Problem Statement:
Find a general solution. Check your answer by substitution.

Solution:
The characteristic equation is given by:


 * {| style = "width: 100%"

! <p style = "text-align:left"> $$ \ r^{2} +2\pi +\pi^{2}= 0 $$ !! <p style = "text-align:right"> (Eq. 2)
 * }

or


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! <p style = "text-align:left"> $$ \ (r + \pi)(r+ \pi) = 0 $$ !! <p style = "text-align:right"> (Eq. 3)
 * }

Since this is a repeated root, the general solution is of the form:


 * {| style = "width: 100%"

! <p style = "text-align:left"> $$ \ y(x) = c_1 e^{x} + c_2 x e^{x} $$ !! <p style = "text-align:right"> (Eq. 4)
 * }

A general solution is then


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! <p style = "text-align:left"> $$ \ y(x) = (c_1 + c_2 x) e^{-\pi x}  $$ <p style = "text-align:right"> (Eq. 5)
 * }

Problem 6

Given:


 * {| style = "width: 100%"

! <p style = "text-align:left"> $$ \ 10y'' - 32 y' + 25.6 y =0 $$ !! <p style = "text-align:right"> (Eq. 6)
 * }

Problem Statement:
Find a general solution. Check your answer by substitution.

Solution:
The characteristic equation is given by:


 * {| style = "width: 100%"

! <p style = "text-align:left"> $$ \ 10 r^{2} -32 r + 25.6= 0 $$ !! <p style = "text-align:right"> (Eq. 7)
 * }

the expression can be factored using the quadratic formula since it is of the form:


 * {| style = "width: 100%"

! <p style = "text-align:left"> $$ \ a x^{2} + b x + c = 0$$ !! <p style = "text-align:right"> (Eq. 8)
 * }

the quadratic formula states that the roots of the characteristic equation, x, are:


 * {| style = "width: 100%"

! <p style = "text-align:left"> $$ \ x=\frac{-b \pm \sqrt {b^2-4ac}}{2a}$$ !! <p style = "text-align:right"> (Eq. 9)
 * }

We see from the result of the calculation that the solution to the characteristic equation is a repeated root of


 * {| style = "width: 100%"

! <p style = "text-align:left"> $$ \ x = 1.6 $$ !! <p style = "text-align:right"> (Eq. 10) Since this is again a repeated root, the general solution is of the form:
 * }


 * {| style = "width: 100%"

! <p style = "text-align:left"> $$ \ y(x) = c_1 e^{x} + c_2 x e^{x} $$ !! <p style = "text-align:right"> (Eq. 11)
 * }

A general solution is then


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! <p style = "text-align:left"> $$ \ y(x) = (c_1 + c_2 x) e^{-1.6 x}  $$ <p style = "text-align:right"> (Eq. 12)
 * }

R2.5
Contributed by Charles Chiamchittrong.

Problem Statement:
Kreyszig 2011 p.59 pbs. 16-17

We are asked to find a second-order homogeneous linear ODE of the form


 * {| style = "width: 100%"

! <p style = "text-align:left"> $$ \ y'' + ay' + by= 0 $$ !! <p style = "text-align:right"> (Eq. 1) for a given basis of solutions.
 * }

Problem 16
 * {| style = "width: 100%"

! <p style = "text-align:left"> $$ \ e^{2.6x} \, \ e^{-4.3x} $$ !! <p style = "text-align:right">
 * }

Problem 17
 * {| style = "width: 100%"

! <p style = "text-align:left"> $$ \ e^{-\sqrt{5}x} \, \ xe^{-\sqrt{5}x} $$ !! <p style = "text-align:right">
 * }

Solution:
Problem 16

Examining the given basis, we can see that this is a case of two distinct real roots.

The general solution in this case is:
 * {| style = "width: 100%"

! <p style = "text-align:left"> $$ \ y = c_1e^{\lambda_1x} + c_2e^{\lambda_2x} $$ !! <p style = "text-align:right"> (Eq. 2)
 * }

Substituting the basis of solutions, we obtain:
 * {| style = "width: 100%"

! <p style = "text-align:left"> $$ \ y = c_1e^{2.6x} + c_2e^{-4.3x} $$ !! <p style = "text-align:right"> (Eq. 3)
 * }

Comparing Eq. 2 and Eq. 3 we can determine the roots $$\lambda_1$$ and $$\lambda_2$$:


 * {| style = "width: 100%"

! <p style = "text-align:left"> $$ \ \lambda_1 = 2.6 \, \ \lambda_2 = -4.3 $$ !! <p style = "text-align:right"> (Eq. 4a, 4b)
 * }

Now we can write the characteristic equation:


 * {| style = "width: 100%"

! <p style = "text-align:left"> $$ \ (\lambda - \lambda_1)(\lambda - \lambda_2) = 0 $$ !! <p style = "text-align:right"> (Eq. 5)
 * }


 * {| style = "width: 100%"

! <p style = "text-align:left"> $$ \ (\lambda - 2.6)(\lambda + 4.3) = 0 $$ !! <p style = "text-align:right"> (Eq. 6)
 * }


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! <p style = "text-align:left"> $$ \ \lambda^2 + 1.7\lambda - 11.18 = 0 $$ !! <p style = "text-align:right"> (Eq. 7)
 * }

Finally, we can convert the characteristic equation into the corresponding homogeneous ODE:


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! <p style = "text-align:left"> $$ \ y'' + 1.7y' - 11.18y = 0 $$ <p style = "text-align:right"> (Eq. 8)
 * }

Problem 17

Examining the given basis, we can see that this is a case of a real double root.

The general solution in this case is:
 * {| style = "width: 100%"

! <p style = "text-align:left"> $$ \ y = (c_1 + c_2x)e^{\lambda x} $$ !! <p style = "text-align:right"> (Eq. 9)
 * }

Substituting the basis of solutions, we obtain:
 * {| style = "width: 100%"

! <p style = "text-align:left"> $$ \ y = (c_1 + c_2x)e^{-\sqrt{5} x} $$ !! <p style = "text-align:right"> (Eq. 10)
 * }

Comparing Eq. 8 and Eq. 9 we can determine the root $$\lambda_1$$:


 * {| style = "width: 100%"

! <p style = "text-align:left"> $$ \ \lambda_1 = -\sqrt{5} $$ !! <p style = "text-align:right"> (Eq. 11)
 * }

Now we can write the characteristic equation:


 * {| style = "width: 100%"

! <p style = "text-align:left"> $$ \ (\lambda - \lambda_1)^2 = 0 $$ !! <p style = "text-align:right"> (Eq. 12)
 * }


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! <p style = "text-align:left"> $$ \ (\lambda + \sqrt{5})^2 = 0 $$ !! <p style = "text-align:right"> (Eq. 13)
 * }


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! <p style = "text-align:left"> $$ \ \lambda^2 + 2\sqrt{5}\lambda + 5 = 0 $$ !! <p style = "text-align:right"> (Eq. 14)
 * }

Finally, we can convert the characteristic equation into the corresponding homogeneous ODE:


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! <p style = "text-align:left"> $$ \ y'' + 2\sqrt{5}y' + 5y = 0 $$ <p style = "text-align:right"> (Eq. 15)
 * }

R2.6
Contributed by Santiago Marin.

Problem Statement:
Realize a spring-dashpot-mass systems in series like the one shown in the figure below.

Using the real double root λ = -3, find the values for the parameters k,c,m.

Solution:
Using equation (3) p.5-5 and putting into standard form, the motion of a spring-dashpot-mass system is:

! <p style = "text-align:left"> $$ y'' + \frac{k}{cm}y' + \frac{k}{m}y = f(t)\!$$ !! <p style = "text-align:right"> (Eq. 1)
 * }

Using the real double root λ= -3, the characteristic equation is:
 * {| style = "width: 100%"

! <p style = "text-align:left"> $$ y'' + 6y' + 9 = 0 \!$$ !! <p style = "text-align:right"> (Eq. 2)
 * }

Matching the coefficients of equation (Eq. 1) and (Eq. 2) a solution is:

$$ c = \frac{3}{2} \!$$

$$ k = 9 \!$$

$$ m = 1 \!$$

R2.7
Contributed by Grant (Kyle) Uppercue.

Problem statement:
Develop the MacLaurin series (Taylor series at t = 0) for:
 * {| style = "width: 100%"

! <p style = "text-align:left"> $$ \ e^t, cost , sint $$ !! <p style = "text-align:right"> (Eq. 1)
 * }

Solutions:
General Taylor Series Expansion at point a:
 * {| style = "width: 100%"

! <p style = "text-align:left"> $$ f(a) + \frac{f'(a)} {1!} (x-a) + \frac{f(a)} {2!} (x-a)^2 + \frac{f'(a)} {3!} (x-a)^3 + ... $$ !! <p style = "text-align:right"> (Eq. 2)
 * }

$$ \ e^t $$, where t = 0:
 * {| style = "width: 100%"

! <p style = "text-align:left"> $$ e^0 + \frac{e^0} {1!} (x) + \frac{e^0} {2!} (x)^2 + \frac{e^0} {3!} (x)^3 + ... $$ !! <p style = "text-align:right"> (Eq. 3)
 * }


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! <p style = "text-align:left"> $$ = 1 + x + \frac{x^2} {2} + \frac{x^3} {6} + ... $$ <p style = "text-align:right"> (Eq. 4)
 * }

$$ \ \cos(t) $$, where t = 0:


 * {| style = "width: 100%"

! <p style = "text-align:left"> $$ \cos(0) + \frac{-\sin(0)} {1!} (x) + \frac{-\cos(0)} {2!} (x)^2 + \frac{sin(0)} {3!} (x)^3 + \frac{cos(0)} {4!} (x)^4 + ... $$ !! <p style = "text-align:right"> (Eq. 5)
 * }


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! <p style = "text-align:left"> $$ = 1 - \frac{x^2} {2} + \frac{x^4} {24} + ... $$ <p style = "text-align:right"> (Eq. 6)
 * }

$$ \ \sin(t) $$, where t = 0:


 * {| style = "width: 100%"

! <p style = "text-align:left"> $$ \sin(0) + \frac{cos(0)} {1!} (x) + \frac{-\sin(0)} {2!} (x)^2 + \frac{-cos(0)} {3!} (x)^3 + \frac{\sin(x)} {4!} (x)^4 + \frac{\cos(0)} {5!} (x)^5 + ... $$ !! <p style = "text-align:right"> (Eq. 7)
 * }


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! <p style = "text-align:left"> $$ = x - \frac{x^3} {3!} + \frac{x^5} {5!} + ... $$ <p style = "text-align:right"> (Eq. 8)
 * }

R2.8
Contributed by Vladimir Horwitz.

Problem Statement:
Answer questions 8 and 15 from Kreyszing.

Solution:
Prob. 8)


 * {| style = "width: 100%"

! <p style = "text-align:left"> $$ \ y'' + y' + 3.25 y = 0 $$ !! <p style = "text-align:right"> (Eq. 1)
 * <p style = "text-align:left"> $$ \ \lambda^2 + \lambda + 3.25 = 0 $$
 * <p style = "text-align:right"> (Eq. 2)
 * }
 * }

The characteristic equation has these roots


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 * <p style = "text-align:left"> $$ \lambda = -1/2 + \sqrt {3} i, \bar \lambda = -1/2 - \sqrt {3} i $$
 * <p style = "text-align:right"> (Eq. 3, a and b)
 * }
 * }

Due to the fact that these roots are complex, the following equations from Kreyszig 2011, page 57 case 3, can be applied.


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! <p style = "text-align:left"> $$ \ y = e^{-1/2} * (A \cos (\sqrt 3 x) + B \sin (\sqrt 3 x)) $$ <p style = "text-align:right"> (Eq. 4)
 * }

And following this, we can say


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 * <p style = "text-align:left"> $$ y' = -.5 e^{-.5} \ (A \cos (\sqrt 3 x) + B \sin (\sqrt 3 x)) \ + \ e^{-.5} \ (-A \sqrt 3 \sin (\sqrt 3 x) + B \sqrt 3 \cos (\sqrt 3 x)) $$
 * <p style = "text-align:right">  (Eq. 5)
 * and
 * <p style = "text-align:left"> $$ y'' = .25 e^{-.5} \ (A \cos \sqrt 3 x \ + B \sin \sqrt 3 x) \ + \ e^{-.5} \ (-3 A \cos \sqrt 3 x \ - 3 B \sin \sqrt 3 x) + \downarrow $$
 * <p style = "text-align:left"> $$ \uparrow -.5 e^{-.5} \ (-A \sqrt 3 \sin \sqrt 3 x \ + B \sqrt 3 \cos \sqrt 3 x) \ - \ .5 e^{-.5} \ (-A \sqrt 3 \sin \sqrt 3 x \ + B \sqrt 3 \cos \sqrt 3 x) $$
 * <p style = "text-align:right"> (Eq. 6)
 * }
 * <p style = "text-align:left"> $$ \uparrow -.5 e^{-.5} \ (-A \sqrt 3 \sin \sqrt 3 x \ + B \sqrt 3 \cos \sqrt 3 x) \ - \ .5 e^{-.5} \ (-A \sqrt 3 \sin \sqrt 3 x \ + B \sqrt 3 \cos \sqrt 3 x) $$
 * <p style = "text-align:right"> (Eq. 6)
 * }

which simplifies to


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 * <p style = "text-align:left"> $$ 0 = (3.25 - .5 + .25 - 3) * e^{-.5} * (A \cos \sqrt 3 x \ + B \sin \sqrt 3 x) $$
 * <p style = "text-align:right"> (Eq. 7)
 * }

Prob. 15)


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! <p style = "text-align:left"> $$ \ y'' + .54 y' + (.0729 + \pi) y = 0 $$ !! <p style = "text-align:right"> (Eq. 1)
 * <p style = "text-align:left"> $$ \ \lambda^2 + .54 \lambda + .0729 + \pi = 0 $$
 * <p style = "text-align:right"> (Eq. 2)
 * }
 * }

The characteristic equation has these roots


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 * <p style = "text-align:left"> $$ \lambda = -.27 + \sqrt {\pi} i, \bar \lambda = -.27 - \sqrt {\pi} i $$
 * <p style = "text-align:right"> (Eq. 3, a and b)
 * }

Due to the fact that this roots are complex, the following equations from Kreyszig 2011, page 57 case 3, can be applied.


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! <p style = "text-align:left"> $$ \ y = e^{-.27} * (A \cos (\sqrt \pi x) + B \sin (\sqrt \pi x)) $$ <p style = "text-align:right"> (Eq. 4)
 * }

And then we can show


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 * <p style = "text-align:left"> $$ y' = -.27 e^{-.27} \ (A \cos (\sqrt \pi x) + B \sin (\sqrt \pi x)) \ + \ e^{-.27} \ (-A \sqrt \pi \sin (\sqrt \pi x) + B \sqrt \pi \cos (\sqrt \pi x)) $$
 * <p style = "text-align:right"> (Eq. 5)
 * and
 * <p style = "text-align:left"> $$ y'' = .0729 e^{-.27} \ (A \cos \sqrt \pi x \ + B \sin \sqrt \pi x) \ + \ e^{-.27} \ (- \pi A \cos \sqrt \pi x \ - \pi B \sin \sqrt \pi x) + \downarrow $$
 * <p style = "text-align:left"> $$ \uparrow -.27 e^{-.27} \ (-A \sqrt \pi \sin \sqrt \pi x \ + B \sqrt \pi \cos \sqrt \pi x) \ - \ .27 e^{-.27} \ (-A \sqrt \pi \sin \sqrt \pi x \ + B \sqrt \pi \cos \sqrt \pi x) $$
 * <p style = "text-align:right"> (Eq. 6)
 * Then, after major calculator work, we find that
 * <p style = "text-align:left"> $$ 0 = (0) * e^{-.27} * (A \cos \sqrt \pi x \ + B \sin \sqrt \pi x) $$
 * <p style = "text-align:right"> (Eq. 7)
 * }
 * Then, after major calculator work, we find that
 * <p style = "text-align:left"> $$ 0 = (0) * e^{-.27} * (A \cos \sqrt \pi x \ + B \sin \sqrt \pi x) $$
 * <p style = "text-align:right"> (Eq. 7)
 * }
 * }

R2.9
Contributed by Gabriel Arab.

Problem Statement:
Find and plot the solution for the L2-ODE-CC,


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! <p style = "text-align:left"> $$ \ \lambda^2 + 4 \lambda + 13 = 0 $$ !! <p style = "text-align:right"> (Eq. 1)
 * Initial conditions,
 * <p style = "text-align:left"> $$ \ y (0) = 1, \ y' (0) = 0 $$
 * <p style = "text-align:right"> (Eq. 2)
 * No excitation,
 * <p style = "text-align:left"> $$ \ r (x) = 0 $$
 * <p style = "text-align:right"> (Eq. 3)
 * }
 * <p style = "text-align:left"> $$ \ r (x) = 0 $$
 * <p style = "text-align:right"> (Eq. 3)
 * }
 * }

Solution:
Solving for roots using Quadratic Equation with a=1, b=4 and c=13


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!<p style = "text-align:left"> $$ \ \lambda = {-b \pm \sqrt {b^2 - 4 a c} \over 2 a} $$ !! <p style = "text-align:right"> (Eq. 4)
 * <p style = "text-align:left"> $$ \ \lambda = {-4 \pm \sqrt {4^2 - 4 * 1 * 13} \over 2 * 1} $$
 * <p style = "text-align:right"> (Eq. 5)
 * which is complex
 * <p style = "text-align:left"> $$ \ \lambda = -2 \pm 3 i $$
 * <p style = "text-align:right"> (Eq. 6)
 * }
 * <p style = "text-align:left"> $$ \ \lambda = -2 \pm 3 i $$
 * <p style = "text-align:right"> (Eq. 6)
 * }

Implementing the solution equation for complex roots and deriving


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! <p style = "text-align:left"> $$ \ y_h (x) = e^{-2 x} (A \cos(3 x) + B \sin(3 x)) $$ !! <p style = "text-align:right"> (Eq. 7)
 * <p style = "text-align:left"> $$ \ y_h' (x) = -2 e^{-2 x} [A \cos(3 x) + B \sin(3 x)] + e^{-2 x} [-3 A \sin(3 x) + 3 B \cos(3 x)] $$
 * <p style = "text-align:right"> (Eq. 8)
 * }
 * }

Plugging in initial conditions into (7) and (8),


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! <p style = "text-align:left"> $$ \ y_h (0) = e^{-2 (0)} [A \cos(3(0)) + B \sin(3(0))] = 1 $$ !! <p style = "text-align:right"> (Eq. 9)
 * <p style = "text-align:left"> $$ \ A = 1 $$
 * <p style = "text-align:right"> (Eq. 10)
 * and
 * <p style = "text-align:left"> $$ \ y_h' (0) = -2 e^{-2 (0)} [A \cos(3 (0)) + B \sin(3 (0))] + e^{-2 (0)} [-3 A \sin(3 (0)) + 3 B \cos(3 (0))] = 0 $$
 * <p style = "text-align:right"> (Eq. 11)
 * <p style = "text-align:left"> $$ \ 2 A + 3 B = 0 $$
 * <p style = "text-align:right"> (Eq. 12)
 * Thus,
 * <p style = "text-align:left"> $$ \ B = 2 / 3 $$
 * <p style = "text-align:right"> (Eq. 13)
 * }
 * Thus,
 * <p style = "text-align:left"> $$ \ B = 2 / 3 $$
 * <p style = "text-align:right"> (Eq. 13)
 * }
 * <p style = "text-align:right"> (Eq. 13)
 * }

Writing final solution equation and plotting on WolframAlpha engine,


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! <p style = "text-align:left"> $$ \ y_h (x) = e^{-2 x} (\cos(3 x) + {2 \over 3} \sin(3 x)) $$ <p style = "text-align:right"> (Eq. 14)
 * }

Page information
Page designed by Vladimir Horwitz on 02:31, 3 February 2012 (UTC).

Last edited by Egm4313.s12.team19.Horwitz on 19:51, 8 February 2012 (UTC).