University of Florida/Egm4313/s12.team12.R2

Report 2

Given

$\lambda _{1}=-2,\lambda _{2}=+5$

$y(0)=1,y'(0)=0$

For part one find the non-homogeneous L2-ODE-CC in standard form and the solution in terms of the initial conditions and the general excitation $r(x).$

Consider no excitation and plot the solution

$r(x)=0$

For part two you need to generate 3 non-standard (and non-homogeneous) L2-ODE-CC that admit the two values $\lambda _{1}=-2,\lambda _{2}=+5\$ as the two roots of the corresponding characteristic equation.

Solution

Find characteristic equation

$(\lambda -(-2))(\lambda -(5))=0$

$\lambda ^{2}-3\lambda -10=0$

The homogeneous solution takes the form

$y_{h}=C_{1}e^{-}2t+C_{2}e^{5}t$

Take the derivative and subitute the intial conditions into the equations

$y'_{h}=-2C_{1}e^{-}2t+5C_{2}e^{5}t$

$y(0)>>1=C_{1}+C_{2}$

$y'(0)>>0=-2C_{1}+5C_{2}$

$C_{2}=1-C_{1}$

$0=-2C_{1}+5-5C_{1}$

$-5=-7C_{1}$

$C_{1}=5/7$

$C_{2}=2/7$

The Solution to the non homogeneous L2-ODE

$y(t)=5/7C_{1}e^{-}2t+2/7C_{2}e^{5}t+y_{p}$

Part two:

Characteristic Equation

$(\lambda -3)(\lambda -(-6))=0$ $\lambda ^{2}+3\lambda -18$

Multiply equation by a variable, i chose m = 2,4,6 respectfully

$2\lambda ^{2}+6\lambda -36$ $4\lambda ^{2}+12\lambda -72$ $6\lambda ^{2}+18\lambda -108$

Author

Patrick Greivell, 15:45 8 February 2012 (UTC)

Problem 2.2

Given

Given is an Initial Value Problem (IVP) to solve a second-order Ordinary Differential Equation (ODE) by finding a particular solution to the equation and its subsequent plot.

Initial Values:

$y(0)=1$

$y'(0)=0$

Differential Equation:

$y''-10y'+25y=r(x)$

Assume that no excitation exists, i.e.

$r(x)=0$

Therefore the ODE to be solved becomes:

$y''-10y'+25y=0$

Solution

With the given conditions, a homogenous linear second order ODE with constant coefficients is given to solve. In order to solve this ODE, a basis of solutions must be formed that meet the given conditions. Our goal is to find a solution to that will fit to the a general solution in the form of:

$y=e^{\lambda x}$

To do this, a quadratic equation in the form of the following is used to determine the method of how the basis of solutions will be formed.

$\lambda ^{2}+a\lambda +b=0$

In order to do find the general solution, and ultimately the particular solution for this IVP, we will need to solve for λ using the characteristic equation above. Given that this is a quadratic equation, we can solve for λ by finding the roots of the equation. We know the variables a and b from our given ODE.

Let:

a = -10 b = 25

Plugging this into the characteristic equation, we yield the following:

$\lambda ^{2}-10\lambda +25=0$

Now, instead of solving directly for λ, we can determine exactly how many and what kind of roots the equation will have based on the following cases given the value of the discriminant:

Case 1	Case 2	Case 3
Discriminant > 0	Discriminant = 0	Discriminant < 0
Two Real Roots	One Real Root	Two Complex Roots

The discriminant is given by the following equation:

$a^{2}-4b$

If we evaluate the discriminant using the values of a and b from above, we can evaluate the discriminant to be:

$(-10)^{2}-4(25)=0$

Therefore, we are in Case 2, where the value of our roots will be one real double root. Therefore, we can determine the value of λ by the following equation and by substituting the values into the equation subsequently.

From this, we can use the value of λ to form one equation of the basis as follows:

In order to form a basis, a second solution is needed. This is done by performing a reduction of order, yielding a first order differential equation. This process will produce the following general solution*:

Substituting the value for λ that was found:

In order to solve for the variables c₁ and c₂, we will use the initial conditions that were given in the problem statement. Therefore, we have two unknowns. We can solve for both of these variables by writing two equations, since one of the initial values was given as a value for the first derivative of the solution at a given point. The derivative of the general solution obtained (shown directly above) is obtained:

We can determine the value of c₂ in terms of c₁ by simply substituting in the known initial values as follows:

The second equation to solve for the unknown variable c₁ can be determined by simply using the general solution equation and using the other initial value in the problem statement.

Therefore, we now know both of the unknown variables. Therefore the particular solution of the given ODE is:

It's subsequent plot is therefore:

Author

--Egm4313.s12.team12.stewart 04:35, 6 February 2012 (UTC)

• Proof for form of general solution found in: Kreyszig, Erwin, Herbert Kreyszig, and E. J. Norminton. Advanced Engineering Mathematics. Hoboken, NJ: Wiley, 2011. Print.

• Plot for particular solution plotted using Wolfram Alpha

Problem 2.3

Statement

K 2011 p.59 pbs. 3-4
Find a general solution. Check your answer by substitution.

Solution

Problem 3:
Consider an ODE of the form:

$\displaystyle {y}''+6{y}'+8.96y=0$

First we notice that the ODE is homogenous, linear, and has constant coefficients.
Therefore, we find the auxiliary equation, which is of the form:

$\displaystyle \lambda ^{2}+a\lambda +b=0$

The value of the discriminant is:

$\displaystyle 6^{2}-4(8.96)=0.16>0$
We plug in a = 6 and b = 8.96 into the auxillary equation to find the roots.

$\displaystyle \lambda ^{2}+6\lambda +8.96=0,\lambda _{1}=-3.2,\lambda _{2}=-2.8$

The general solution will be of the form:

$\displaystyle y=c_{1}e^{\lambda _{1}x}+c_{2}e^{\lambda _{2}x}$

Therefore the solution will be

$\displaystyle y=c_{1}e^{-3.2x}+c_{2}e^{-2.8x}$

Problem 4:
Consider an ODE of the form:

$\displaystyle {y}''+4y'+(\pi ^{2}+4)y=0$

First we notice that the ODE is homogenous, linear, and has constant coefficients.
Therefore, we find the auxiliary equation, which is of the form:

$\displaystyle \lambda ^{2}+a\lambda +b=0$

Looking at the coefficients we obtain:

$\displaystyle \lambda ^{2}+4\lambda +(\pi ^{2}+4)=0$

Next we obtain the roots of this equation with the quadratic formula:

$\displaystyle \ {\frac {-4\pm {\sqrt {(-4)^{2}-4(1)(\pi ^{2}+4)}}}{2(1)}}$

This can be simplified to:

$\displaystyle -2\pm {\frac {\sqrt {16-4\pi ^{2}-16}}{2}}$

$\displaystyle -2\pm {\frac {\sqrt {-4\pi ^{2}}}{2}}$

We notice that by looking inside the radical, we will be obtaining complex roots. Continuing with simplification we get:

$\displaystyle -2\pm {i\pi }$

The general solution to the case of complex roots is:

$\displaystyle y=e^{\frac {-ax}{2}}(Acos\omega {x}+Bsin\omega {x})$

Where complex roots are of the form:

$\displaystyle -{\frac {1}{2}}a\pm {i\omega }$

Knowing $\displaystyle a=4$ and $\displaystyle \omega =\pi$, we get the final general solution:

$\displaystyle y=e^{-2x}(Acos\pi {x}+Bsin\pi {x})$

Author

Egm4313.s12.team12.sutcliffe 18:42, 8 February 2012 (UTC)

Problem 2.4

Statement

K 2011 p.59 pbs. 5-6
Find a general solution. Check your answer by substitution.

Solution

Problem 5:
Consider an ODE of the form:

$\displaystyle {y}''+2\pi y'+\pi ^{2}y=0$

This ODE is linear, and has constant coefficients. Therefore, its general solution is of the form:

$\displaystyle {y}=e^{\lambda x}$

Therefore:

$\displaystyle {y}'=\lambda e^{\lambda x}$

And:

$\displaystyle {y}''={\lambda }^{2}e^{\lambda x}$

The ODE can be rearranged so that:

$\displaystyle e^{\lambda x}(\lambda ^{2}+2\pi \lambda +\pi ^{2})=0$

And since:

$\displaystyle e^{\lambda x}\neq 0$

Then:

$\displaystyle \lambda ^{2}+2\pi \lambda +\pi ^{2}=0$

$\displaystyle (\lambda +\pi )=0$

$\displaystyle \lambda =-\pi$

Since:

$\displaystyle \lambda _{1}=\lambda _{2}$

The general solution then becomes:

$\displaystyle y=c_{1}e^{\lambda x}+c_{2}xe^{\lambda x}$

$\displaystyle y=e^{\lambda x}(c_{1}+c_{2}x)$

Therefore:

$\displaystyle y=e^{-\pi x}(c_{1}+c_{2}x)$

Problem 6:
Consider an ODE of the form:

$\displaystyle 10{y}''-32{y}'+25.66y=0$

Divide through by 10 to get the equation into standard form where the leading coefficient is equal to 1.

$\displaystyle {y}''-3.2{y}'+2.56y=0$

First we notice that the ODE is homogenous, linear, and has constant coefficients.
Therefore, we find the auxiliary equation, which is of the form:

$\displaystyle \lambda ^{2}+a\lambda +b=0$

The characteristic equation then becomes

$\displaystyle \lambda ^{2}-3.2\lambda +2.56=0$

The value of the discriminant is:

$\displaystyle (-3.2)^{2}-4(2.56)=0$

The 0 indicates that there is a real double root.

The general solution will be of the form:

$\displaystyle y=e^{-{\frac {a}{2}}x}(C_{1}+C_{2}x)$

Where -a/2 = -(-3.2)/2 = 1.6

Therefore the solution will be

$\displaystyle y=e^{1.6x}(C_{1}+C_{2}x)$

Author

Egm4313.s12.team12.sutcliffe 18:43, 8 February 2012 (UTC)

Problem 2.5

Given

K 2011 p.59 pbs. 16-17
Find an ODE

$\displaystyle y''+ay'+by=0$

for the given basis:

$\displaystyle 16.e^{2.6x},e^{-4.3x}$

$\displaystyle 17.e^{-{\sqrt {5}}x},xe^{-{\sqrt {5}}x}$

Solution

Problem 16.
Distinct real roots are of the form:

$\displaystyle e^{\lambda _{1}x},e^{\lambda _{2}x}$

Therefore, $\displaystyle \lambda _{1}=2.6,\lambda _{2}=-4.3$

The characteristic equation is:

$\displaystyle (\lambda -\lambda _{1})(\lambda -\lambda _{2})=0$

Therefore we have:
$\displaystyle (\lambda -2.6)(\lambda -(-4.3))=0$

Through simplification and distribution we get:
$\displaystyle \lambda ^{2}+1.7\lambda -11.18=0$

This is the auxiliary equation to the original linear, homogeneous ODE.
Therefore our final solution is:

$\displaystyle y''+1.7y'-11.18y=0$

Problem 17.
Double real roots are of the form:

$\displaystyle e^{-ax/2},xe^{-ax/2}$

Therefore, $\displaystyle \lambda =-{\sqrt {5}}$

The characteristic equation is:

$\displaystyle (\lambda -(-{\sqrt {5}}))^{2}=\lambda ^{2}+2{\sqrt {5}}\lambda +5=0$

This is the auxiliary equation to the original linear, homogeneous ODE.
Therefore our final solution is:

$\displaystyle y''+2{\sqrt {5}}y'+5y=0$

Author

Egm4313.s12.team12.stadick 24:00, 7 February 2012 (UTC)

Given

Find the parameters k, c, m of the the spring-dotplot-mass system with a double root $\lambda =-3$

Solution

The equation of motion for this system is

$m(y''_{k}+{\frac {k}{c}}y'_{k})+ky_{k}=f(t)$

or

$my''_{k}+m{\frac {k}{c}}y'_{k}+ky_{k}=f(t)$

The homogeneous equation is

$Ay''_{h}+By'_{h}+Cy_{h}=0$

Characteristic Equation of the double root

$(\lambda -(-3))^{2}=\lambda ^{2}+6\lambda +9$

Then you find the homogeneous L2-ODE-CC for the characteristic eq.

$y''+6y'+9y=0$

Now compare the coeffients with the motion equation

$m=1$

$k=9$

${\frac {mk}{c}}=6$

$c={\frac {1*9}{6}}$

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

Author

Patrick Greivell, 16:16 8 February 2012(UTC)

Problem 2.7

Given

Develop the Maclaurin Series (Taylor Series @ t = 0) for

a) $e^{t}$

b) $cos(t)$

c) $sin(t)$

Solution

The General form of a Taylor Series is

$\sum _{n=0}^{\infty }{\frac {f^{n}(a)}{n!}}(t-a)^{n}$

with the expansion at point a being

$f(a)+{\frac {f'(a)}{1!}}(x-a)+{\frac {f''(a)}{2!}}(x-a)^{2}+{\frac {f^{3}(a)}{3!}}(x-a)^{3}+...$

The Maclaurin Series (Taylor Series @ t=0) would be

a)

$e^{t}=\sum _{n=0}^{\infty }{\frac {t^{n}}{n!}}$

which would equal

$e^{0}+{\frac {e^{0}}{1!}}(x)+{\frac {e^{0}}{2!}}(x)^{2}+{\frac {e^{0}}{3!}}(x)^{3}+...$

or

$1+t+{\frac {x^{2}}{2}}+{\frac {x^{3}}{6}}+...$ for all x

b)

$cos(t)=\sum _{n=0}^{\infty }{\frac {(-1)^{n}t^{2n}}{(2n)!}}$

which would equal

$cos(0)+{\frac {-sin(0)}{1!}}(x)+{\frac {-cos(0)}{2!}}(x)^{2}+{\frac {sin(0)}{3!}}(x)^{3}+{\frac {cos(0)}{4!}}(x)^{4}+...$

or

$1+0+{\frac {x^{2}}{2}}+0+{\frac {x^{4}}{24}}+...$ for all x

c)

$sin(t)=\sum _{n=0}^{\infty }{\frac {(-1)^{n}t^{2n+1}}{(2n+1)!}}$

which would equal

$sin(0)+{\frac {cos(0)}{1!}}(x)+{\frac {-sin(0)}{2!}}(x)^{2}+{\frac {-cos(0)}{3!}}(x)^{3}+{\frac {sin(0)}{4!}}(x)^{4}+...$

or

$0+x+0+{\frac {x^{3}}{6}}+0+...$ for all x

Author

Cpettigrew 21:15, 5 February 2012 (UTC)

Problem 2.8 (a)

Given

#8 P. 59 Given $\displaystyle y''+y'+3.25y=0$ find a general solution and check your answer by substitution.

Solution

$\displaystyle 1y''+1y'+3.25y=0$

$\displaystyle a^{2}-4b=(1)^{2}-(4)(1)=1-4=-3$

Since $\displaystyle a^{2}-4b<0$ it is a case where the solutions are complex conjugate roots.

A general solution for an ODE of this case can be given by the equation $\displaystyle y=e^{-ax/2}(A\cos \omega x+B\sin \omega x)$

1st Step

The 1st step to finding the solutions for this ODE is to substitute the variable $\displaystyle \lambda$ in for y and its derivatives.

$\displaystyle \lambda ={\frac {dy}{dx}}$ and $\displaystyle \lambda ^{2}={\frac {d^{2}y}{dx^{2}}}$

So $\displaystyle y''+y'+3.25y=0$ becomes $\displaystyle \lambda ^{2}+\lambda +3.25=0$

2nd Step

Now that the ODE has been made into a quadratic equation of form $\displaystyle ax^{2}+bx+c=0$ we can use the quadratic formula $\displaystyle x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}$

Inputting the coefficients from the equation and solving for $\displaystyle \lambda$ you get $\displaystyle \lambda ={\frac {-1\pm i{\sqrt {12}}}{2}}$

or $\displaystyle \lambda _{1}={\frac {-1+i{\sqrt {12}}}{2}}$ and $\displaystyle \lambda _{2}={\frac {-1-i{\sqrt {12}}}{2}}$

where $\displaystyle \lambda _{1}=-{\frac {1}{2}}a+i\omega$ and $\displaystyle \lambda _{2}=-{\frac {1}{2}}a-i\omega$

so $\displaystyle \omega ={\frac {\sqrt {12}}{2}}$ or $\displaystyle \omega ={\sqrt {3}}$

3rd Step

Now if you plug all of this in to the general solution you will get $\displaystyle y=e^{-{\frac {1}{2}}x}(A\cos {\sqrt {3}}x+B\sin {\sqrt {3}}x)$

Checking Solution with Substitution

Using the general solution and taking the first and second derivatives of this general solution we can plug that in to the original equation and see if we in fact get zero. If so, then the general solution that was found is a solution for the ODE.

$\displaystyle y=e^{-{\frac {1}{2}}x}(A\cos {\sqrt {3}}x+B\sin {\sqrt {3}}x)$

The first derivative is $\displaystyle y'=-{\frac {1}{2}}e^{-{\frac {1}{2}}x}(A\cos {\sqrt {3}}x+B\sin {\sqrt {3}}x)+e^{-{\frac {1}{2}}x}(-{\sqrt {3}}A\sin {\sqrt {3}}x+{\sqrt {3}}B\cos {\sqrt {3}}x)$

The second derivative is $\displaystyle y''={\frac {1}{4}}e^{-{\frac {1}{2}}x}(A\cos {\sqrt {3}}x+B\sin {\sqrt {3}}x)-{\frac {1}{2}}e^{-{\frac {1}{2}}x}(-{\sqrt {3}}A\sin {\sqrt {3}}x+{\sqrt {3}}B\cos {\sqrt {3}}x)-{\frac {1}{2}}e^{-{\frac {1}{2}}x}(-{\sqrt {3}}A\sin {\sqrt {3}}x+{\sqrt {3}}B\cos {\sqrt {3}}x)$

$\displaystyle +e^{-{\frac {1}{2}}x}(-3A\cos {\sqrt {3}}x-3B\sin {\sqrt {3}}x)$

Plugging all of this into the original equation you get

$\displaystyle {\frac {1}{4}}e^{-{\frac {1}{2}}x}(A\cos {\sqrt {3}}x+B\sin {\sqrt {3}}x)-{\frac {1}{2}}e^{-{\frac {1}{2}}x}(-{\sqrt {3}}A\sin {\sqrt {3}}x+{\sqrt {3}}B\cos {\sqrt {3}}x)-{\frac {1}{2}}e^{-{\frac {1}{2}}x}(-{\sqrt {3}}A\sin {\sqrt {3}}x+{\sqrt {3}}B\cos {\sqrt {3}}x)$

$\displaystyle +e^{-{\frac {1}{2}}x}(-3A\cos {\sqrt {3}}x-3B\sin {\sqrt {3}}x)-{\frac {1}{2}}e^{-{\frac {1}{2}}x}(A\cos {\sqrt {3}}x+B\sin {\sqrt {3}}x)+e^{-{\frac {1}{2}}x}(-{\sqrt {3}}A\sin {\sqrt {3}}x+{\sqrt {3}}B\cos {\sqrt {3}}x)$

$\displaystyle +3.25e^{-{\frac {1}{2}}x}(A\cos {\sqrt {3}}x+B\sin {\sqrt {3}}x)=0$

Simplifying all of these terms you end up getting

$\displaystyle e^{-{\frac {1}{2}}x}(0)=0$

Author

--Egm4313.s12.team12.anders. 4:11, 7 February 2012 (UTC)

Problem 2.8 (b)

Given

#15 P.59 Given $\displaystyle y''+0.54y'+(0.0729+\pi )y=0$ find a general solution and check your answer by substitution.

Solution

$\displaystyle 1y''+0.54y'+(0.0729+\pi )y=0$

$\displaystyle a^{2}-4b=(1)^{2}-(4)(0.54)=1-2.16=-1.16$

Since $\displaystyle a^{2}-4b<0$ it is a case where the solutions are complex conjugate roots.

A general solution for an ODE of this case can be given by the equation $\displaystyle y=e^{-ax/2}(A\cos \omega x+B\sin \omega x)$

1st Step

The 1st step to finding the solutions for this ODE is to substitute the variable $\displaystyle \lambda$ in for y and its derivatives.

$\displaystyle \lambda ={\frac {dy}{dx}}$ and $\displaystyle \lambda ^{2}={\frac {d^{2}y}{dx^{2}}}$

So $\displaystyle y''+0.54y'+(0.0729+\pi )y=0$ becomes $\displaystyle \lambda ^{2}+0.54\lambda +(0.0729+\pi )=0$

2nd Step

Now that the ODE has been made into a quadratic equation of form $\displaystyle ax^{2}+bx+c=0$ we can use the quadratic formula $\displaystyle x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}$

Inputting the coefficients from the equation and solving for $\displaystyle \lambda$ you get $\displaystyle \lambda ={\frac {-1\pm i{\sqrt {12.56637061}}}{2}}$

or $\displaystyle \lambda _{1}={\frac {-1+i{\sqrt {12.56637061}}}{2}}$ and $\displaystyle \lambda _{2}={\frac {-1-i{\sqrt {12.56637061}}}{2}}$

where $\displaystyle \lambda _{1}=-{\frac {1}{2}}a+i\omega$ and $\displaystyle \lambda _{2}=-{\frac {1}{2}}a-i\omega$

so $\displaystyle \omega ={\frac {\sqrt {12.56637061}}{2}}$ or $\displaystyle \omega ={\sqrt {\pi }}$

3rd Step

Now if you plug all of this in to the general solution you will get $\displaystyle y=e^{-0.27x}(A\cos {\sqrt {\pi }}x+B\sin {\sqrt {\pi }}x)$

Checking Solution with Substitution

Using the general solution and taking the first and second derivatives of this general solution we can plug that in to the original equation and see if we in fact get zero. If so, then the general solution that was found is a solution for the ODE.

$\displaystyle y=e^{-0.27x}(A\cos {\sqrt {\pi }}x+B\sin {\sqrt {\pi }}x)$

The first derivative is $\displaystyle y'=-0.27e^{-0.27x}(A\cos {\sqrt {\pi }}x+B\sin {\sqrt {\pi }}x)+e^{-0.27x}(-{\sqrt {\pi }}A\sin {\sqrt {\pi }}x+{\sqrt {\pi }}B\cos {\sqrt {\pi }}x)$

The second derivative is $\displaystyle y''=0.0729e^{-0.27x}(A\cos {\sqrt {\pi }}x+B\sin {\sqrt {\pi }}x)-0.27e^{-0.27x}(-{\sqrt {\pi }}A\sin {\sqrt {\pi }}x+{\sqrt {\pi }}B\cos {\sqrt {\pi }}x)-0.27e^{-0.27x}(-{\sqrt {\pi }}A\sin {\sqrt {\pi }}x+{\sqrt {\pi }}B\cos {\sqrt {\pi }}x)$

$\displaystyle +e^{-0.27x}(-\pi A\cos {\sqrt {\pi }}x-\pi B\sin {\sqrt {\pi }}x)$

Plugging all of this into the original equation you get

$\displaystyle 0.0729e^{-0.27x}(A\cos {\sqrt {\pi }}x+B\sin {\sqrt {\pi }}x)-0.27e^{-0.27x}(-{\sqrt {\pi }}A\sin {\sqrt {\pi }}x+{\sqrt {\pi }}B\cos {\sqrt {\pi }}x)-0.27e^{-0.27x}(-{\sqrt {\pi }}A\sin {\sqrt {\pi }}x+{\sqrt {\pi }}B\cos {\sqrt {\pi }}x)$

$\displaystyle +e^{-0.27x}(-\pi A\cos {\sqrt {\pi }}x-\pi B\sin {\sqrt {\pi }}x)-0.1458e^{-0.27x}(A\cos {\sqrt {\pi }}x+B\sin {\sqrt {\pi }}x)+0.54e^{-0.27x}(-{\sqrt {\pi }}A\sin {\sqrt {\pi }}x+{\sqrt {\pi }}B\cos {\sqrt {\pi }}x)$

$\displaystyle +(0.0729+\pi )e^{-0.27x}(A\cos {\sqrt {\pi }}x+B\sin {\sqrt {\pi }}x)=0$

Simplifying all of these terms you end up getting

$\displaystyle e^{-0.27x}(0)=0$

Author

--Egm4313.s12.team12.anders. 4:11, 7 February 2012 (UTC)

Problem 2.9

Given

The ODE is:

$\displaystyle y''+4y'+13y=r(x)$

The given initial conditions are:

$\displaystyle y(0)=1$

$\displaystyle y'(0)=0$

$\displaystyle r(x)=0$

The corresponding characteristic equation is:

$\displaystyle \lambda ^{2}+4\lambda +13=0$

Solution

Solving for $\lambda$:

$\displaystyle \lambda ={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}$

$\displaystyle \lambda ={\frac {-4\pm {\sqrt {4^{2}-4(1)(13)}}}{2(1)}}$

$\displaystyle \lambda ={\frac {-4\pm {\sqrt {-36}}}{2}}$

$\displaystyle \lambda ={\frac {-4\pm 2{\sqrt {-9}}}{2}}$

$\displaystyle \lambda ={-2\pm 3i}$

Therefore, the general solution will be:

$\displaystyle y=C_{1}e^{a_{1}x}cosb_{1}x+C_{2}e^{a_{2}x}sinb_{2}x$

Where:

$\displaystyle a_{1}=a_{2}=-2$

And:

$\displaystyle b_{1}=b_{2}=3$

Thus:

$\displaystyle y=C_{1}e^{-2x}cos3x+C_{2}e^{-2x}sin3x$

Therefore:

$\displaystyle y(0)=1=C_{1}e^{-2(0)}cos(3(0))+C_{2}e^{-2(0)}sin(3(0))=C_{1}(1)(1)+C_{2}(1)(0)=C_{1}$

$\displaystyle C_{1}=1$

$\displaystyle y'(0)=0=-2C_{1}e^{-2x}cos(3x)-C_{1}e^{-2x}sin(3x)-2C_{2}e^{-2x}sin(3x)+C_{2}e^{-2x}cos(3x)$

$\displaystyle 0=-2C_{1}+3C_{2}$

$\displaystyle C_{2}={\frac {2}{3}}$

Final Solution is as follows:



$\displaystyle \ y=e^{-2x}cos(3x)+{\frac {2}{3}}e^{-2x}sin(3x)\$

Author

Wagner Schulz 13:00, 8 February 2012 (UTC)

Contribution Table

Problem Number	Assigned To	Solution By	Proofread By
2.1	Example	Example	Example
2.2	Daniel Stewart	Daniel Stewart	James Stadick
2.3	Garrett Sutcliffe	Garrett Sutcliffe	Example
2.4	Garrett Sutcliffe	Garrett Sutcliffe	Example
2.5	James Stadick	James Stadick	Example
2.6	Example	Example	Example
2.7	Craig Pettigrew	Craig Pettigrew	Example
2.8	Lauren Anders	Lauren Anders	Example
2.9	Wagner Schulz	Wagner Schulz	Example