User:Egm4313.s12.team4/Report 2

=Problem 1=

Problem Statement
Given the two roots:

$$\displaystyle \lambda_1 = -2 \! $$

$$\displaystyle \lambda_2 = 5 \! $$

and initial conditions:

$$\displaystyle y(0) = 1 \!$$

$$\displaystyle y'(0) = 0 \!$$

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 $$\displaystyle r(x) = 0 \! $$


 * Plot the solution

Solution
Characteristic Equation:

$$(\lambda-\lambda_1)(\lambda-\lambda_2) \!$$

Using the Given values for $$ \lambda_1 \! $$ and $$ \lambda_2 \! $$ we can write:

$$ (\lambda+2)(\lambda-5) = 0 \rightarrow \lambda^2-3\lambda-10 = 0 $$

Non-homogeneous L2-ODE-CC:

$$ y''-3y'-10y = r(x) \! $$

The homogeneous solution is given by:

$$ y_h = c_1 e^{-2x} + c_2 e^{5x} \! $$

The overall solution is given by:

$$ y(x) = y_h + y_p \! $$; where $$ y_h $$ is the homogeneous solution and $$ y_p $$ is the particular solution.

When the excitation factor $$ r(x) \! $$ is zero (as in this case), $$ y_p $$ is zero, so we can use the given initial conditions to find constants $$ c_1 \!$$ and $$ c_2 \!$$.

To do this we must first take the derivative $$ \frac{dy}{dx} $$ of $$ y_h \!$$.

Doing so, we find $$ y'_h = 0 = -2c_1+5c_2 \!$$

Multiplying $$ y_h \!$$ by 2 and adding to $$ y'_h \!$$ we can solve for $$ c_2 \!$$

$$ c_2 = \frac{2}{7} \!$$

Then multiplying $$ y_h $$ by -5 and adding to $$ y'_h $$ we can solve for $$ c_1 $$

$$ c_1 = \frac{5}{7} \!$$

Substituting in our new values for $$ c_1 \!$$ and $$ c_2 \!$$ we find the final solution to be:


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


 * style="width:95%" |
 * style="width:95%" |
 * (2.2.3)
 * }

To obtain three non-standard (non homogeneous) L2-ODE-CC that correspond to the same roots of our characteristic equation, we can simply multiply our characteristic equation by any non-zero constant. So choosing 2,3 and 4 as our constants we obtain:


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


 * style="width:95%" |
 * style="width:95%" |
 * (2.2.3)
 * }


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


 * style="width:95%" |
 * style="width:95%" |
 * (2.2.3)
 * }


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


 * style="width:95%" |
 * style="width:95%" |
 * (2.2.3)
 * }

These will all yield the same roots; $$ \lambda_1 \! $$ and $$ \lambda_2 \! $$

Author
Solved and typed by - Egm4313.sp12.team4.diroma 04:23, 8 February 2012 (UTC) Reviewed By - Team 4

Edited by - Egm4313.sp12.team4.diroma 04:23, 8 February 2012 (UTC)

=Problem 2=

Problem Statement
Find and plot the solution for L2-ODE-CC (2.2.1) with the given initial conditions (2.2.2, 2.2.3) and excitation function (2.2.4).


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

$$\displaystyle y'' - 10y' + 25y = r(x)$$
 * style="width:95%" |
 * style="width:95%" |
 * (2.2.1)
 * }

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

$$\displaystyle y(0) = 1$$
 * style="width:95%" |
 * style="width:95%" |
 * (2.2.2)
 * }


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

$$\displaystyle y'(0) = 0$$
 * style="width:95%" |
 * style="width:95%" |
 * (2.2.3)
 * }

No excitation:
 * {| style="width:100%" border="0"

$$\displaystyle r(x) = 0$$
 * style="width:95%" |
 * style="width:95%" |
 * (2.2.4)
 * }

Solution
Using the following:


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

$$\displaystyle {y}'=\frac{dy}{dx}$$
 * style="width:95%" |
 * style="width:95%" |
 * (2.2.5)
 * }


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

$$\displaystyle {y}''=\frac{d^{2}y}{dx^{2}}$$
 * style="width:95%" |
 * style="width:95%" |
 * (2.2.6)
 * }

The given equation can be rewritten as follows, using (2.2.5) and (2.2.6):
 * {| style="width:100%" border="0"

$$\displaystyle \frac{d^{2}y}{dx^{2}}-10\frac{dy}{dx}+25y=0$$
 * style="width:95%" |
 * style="width:95%" |
 * (2.2.7)
 * }

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

$$\displaystyle D= \frac{d}{dx}$$
 * style="width:95%" |
 * style="width:95%" |
 * (2.2.8)
 * }

The ODE can be expressed as follows:
 * {| style="width:100%" border="0"

$$\displaystyle (D^{2}-10 D+25)y=0$$
 * style="width:95%" |
 * style="width:95%" |
 * (2.2.9)
 * }

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

$$\displaystyle D^{2}-10 D+25=0$$
 * style="width:95%" |
 * style="width:95%" |
 * (2.2.10)
 * }


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

$$\displaystyle (D-5)^{2}=0$$
 * style="width:95%" |
 * style="width:95%" |
 * (2.2.11)
 * }


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

$$\displaystyle D=5,5$$
 * style="width:95%" |
 * style="width:95%" |
 * (2.2.12)
 * }

The general solution for this problem is :


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

$$\displaystyle y=(c_1 + c_2 x) e^{5 x}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.2.13)
 * }

Use initial condition (2.2.2) to find the specific solution,
 * {| style="width:100%" border="0"

$$\displaystyle y(0) = 1 = (c_1 + c_2 (0)) e^0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.2.14)
 * }


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

$$\displaystyle c_1 = 1$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.2.15)
 * }

Find the derivative to apply the second initial condition (2.2.3).
 * {| style="width:100%" border="0"

$$\displaystyle y' = 5 c_1 e^{5 x} + 5 c_2 x e^{5 x} + c_2 e^{5 x}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.2.16)
 * }


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

$$\displaystyle y'(0) = 0 = 5 (1) e^0 + 5 c_2 (0) e^0 + c_2 e^0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.2.17)
 * }


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

$$\displaystyle 0 = 5 + c_2$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.2.18)
 * }


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

$$\displaystyle c_2 = -5$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.2.19)
 * }

The solution to the initial value problem:
 * {| style="width:100%" border="0"


 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.2.20)
 * }

Plot of the solution:
 * {| style="width:100%" border="0"


 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.2.21)
 * }

Author
Solved and typed by - Egm4313.s12.team4.friedman 02:00, 8 February 2012 (UTC) Reviewed By - Team 4

Edited by - Egm4313.s12.team4.friedman 02:00, 8 February 2012 (UTC)

= Problem 3: General Solution for ODEs =

Problem Statement
Problem found in the textbook on page 59, problems 3 and 4. Find a general solution. Check your answer by substitution.


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

a) $$\displaystyle {y}''+6 {y}'+8.96y=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.3a.1)
 * }


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

b) $$\displaystyle {y}''+4{y}'+(\pi^{2}+4)y=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.3b.1)
 * }

Solution (2.3a)
Using the following:


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

$$\displaystyle {y}'=\frac{dy}{dx}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.3a.2)
 * }


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

$$\displaystyle {y}''=\frac{d^{2}y}{dx^{2}}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.3a.3)
 * }

The given equation can be rewritten as follows, using (2.3a.2) and (2.3a.3):
 * {| style="width:100%" border="0"

$$\displaystyle \frac{d^{2}y}{dx^{2}}+6\frac{dy}{dx}+8.96y=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.3a.4)
 * }

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

$$\displaystyle D= \frac{d}{dx}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.3a.5)
 * }

The ODE can be expressed as follows:
 * {| style="width:100%" border="0"

$$\displaystyle (D^{2}+6D+8.96)y=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.3a.6)
 * }

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

$$\displaystyle (D^{2}+6D+8.96)=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.3a.7)
 * }

Use the quadratic formula


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

$$\displaystyle \frac{6\pm\sqrt{6^{2}-4(1)(8.96)}}{2(1)} $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.3a.8)
 * }

Arriving at


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

$$\displaystyle (D+3.2)(D+2.8)=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.3a.9)
 * }


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

$$\displaystyle D=-3.2,-2.8$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.3a.10)
 * }

The final general solution for this problem is:

Reach the general solution :


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


 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.3a.11)
 * }

Check by substitution:

From (2.3a.11), we find,
 * {| style="width:100%" border="0"

$$\displaystyle {y}'=-3.2C_{1}e^{-3.2x}-2.8C_{2}e^{-2.8x}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.3a.12)
 * }


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

$$\displaystyle {y}''=10.24C_{1}e^{-3.2x}+7.84C_{2}e^{-2.8x}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.3a.13)
 * }

Substituting (2.3a.11), (2.3a.12), and (2.3a.13) in (2.3a.1), we reach:
 * {| style="width:100%" border="0"

$$\displaystyle (10.24C_{1}e^{-3.2x}+7.84C_{2}e^{-2.8x})+6(-3.2C_{1}e^{-3.2x}-2.8C_{2}e^{-2.8x})+8.96(C_{1}e^{-3.2x}+C_{2}e^{-2.8x})=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.3a.14)
 * }


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

$$\displaystyle 19.2C_{1}e^{-3.2x}+16.8C_{2}e^{-2.8x} -19.2C_{1}e^{-3.2x}-16.8C_{2}e^{-2.8x}=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.3a.15)
 * }


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

$$\displaystyle 0=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.3a.16)
 * }

Therefore the solution is correct.

Author
Solved and typed by - Egm4313.s12.team4.blj 20:31, 6 February 2012 (UTC) Reviewed By - Team 4

Edited by - Egm4313.s12.team4.blj 20:31, 6 February 2012 (UTC)

Solution (2.3b)
Using the following:


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

$$\displaystyle {y}'=\frac{dy}{dx}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.3b.2)
 * }


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

$$\displaystyle {y}''=\frac{d^{2}y}{dx^{2}}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.3b.3)
 * }

The given equation can be rewritten as follows, using (2.3b.2) and (2.3b.3):
 * {| style="width:100%" border="0"

$$\displaystyle \frac{d^{2}y}{dx^{2}}+4\frac{dy}{dx}+(\pi^2 +4)y=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.3b.4)
 * }

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

$$\displaystyle D= \frac{d}{dx}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.3b.5)
 * }

The ODE can be expressed as follows:
 * {| style="width:100%" border="0"

$$\displaystyle (D^{2}+4D+(\pi^2 +4))y=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.3b.6)
 * }

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

$$\displaystyle (D^{2}+4D+(\pi^2 +4))=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.3b.7)
 * }

Use the quadratic formula


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

$$\displaystyle \frac{4\pm\sqrt{4^{2}-4(1)(\pi^2 +4)}}{2(1)} $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.3b.8)
 * }

Arriving at


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

$$\displaystyle (D+(2+\pi i))(D+(2-\pi i)=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.3b.9)
 * }


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

$$\displaystyle D=-(2+\pi i),-(2-\pi i)$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.3b.10)
 * }

The final general solution for this problem is:

Reach the general solution :


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


 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.3b.11)
 * }

Check by substitution:

From (2.3b.11), we find,
 * {| style="width:100%" border="0"

$$\displaystyle {y}'= -2(C_{1} \cos (\pi x) + C_{2} \sin (\pi x))e^{-2x} + (-C_{1} \pi \sin (\pi x) + C_{2} \pi \cos (\pi x))e^{-2x}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.3b.12)
 * }


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

$$\displaystyle {y}''=4(C_{1} \cos (\pi x) + C_{2} \sin (\pi x))e^{-2x} - 2(-C_{1} \pi \sin (\pi x) + C_{2} \pi \cos (\pi x))e^{-2x}$$ $$\color{White}y''= \color{Black} - 2(-C_{1} \pi \sin (\pi x) + C_{2}\pi \cos (\pi x))e^{-2x} + (-C_{1} \pi ^2 \cos (\pi x) - C_{2} \pi ^2 \sin (\pi x))e^{-2x}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.3b.13)
 * }

Substituting (2.3b.11), (2.3b.12), and (2.3b.13) in (2.3b.1), we reach:
 * {| style="width:100%" border="0"

$$\displaystyle 4(C_{1} \cos (\pi x) + C_{2} \sin (\pi x))e^{-2x} - 2(-C_{1} \pi \sin (\pi x) + C_{2} \pi \cos (\pi x))e^{-2x} - 2(-C_{1} \pi \sin (\pi x) + C_{2} \pi \cos (\pi x))e^{-2x} $$ $$\displaystyle+ (-C_{1}\pi ^2 \cos (\pi x) - C_{2} \pi ^2 \sin (\pi x))e^{-2x} +4 -2(C_{1} \cos (\pi x) + C_{2} \sin (\pi x))e^{-2x}) + (-C_{1} \pi \sin (\pi x) + C_{2} \pi \cos (\pi x))e^{-2x}$$ $$\displaystyle+(\pi^{2}+4)((C_{1}\cos(\pi x) + C_{2}\sin(\pi x))e^{-2x})=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.3b.14)
 * }


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

$$\displaystyle 0=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.3b.15)
 * }

Therefore the solution is correct.

Author
Solved and typed by - Egm4313.s12.team4.blj 20:32, 6 February 2012 (UTC) Reviewed By - Team 4

Edited by - Egm4313.s12.team4.blj 20:32, 6 February 2012 (UTC)

= Problem 4: General Solution for ODEs =

Problem Statement
Problem found in the textbook on page 59, problems 5 and 6. Find a general solution. Check your answer by substitution.


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

5) $$\displaystyle {y}''+2\pi {y}'+\pi^{2}y=0$$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


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

6) $$\displaystyle 10{y}''-32{y}'+25.6=0$$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Solution (Problem 5)
Using the following:


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

$$\displaystyle {y}=e^{rx}$$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


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

$$\displaystyle {y}'=re^{rx}$$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


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

$$\displaystyle {y}''=r^2e^{rx}$$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Now plug in $$\displaystyle {y},{y}',{y}''$$ into the original equation:
 * {| style="width:100%" border="0"

$$\displaystyle r^2e^{rx}+2\pi{r}e^{rx}+\pi^{2}e^{rx}=0$$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Now factor out $$\displaystyle e^{rx}$$:
 * {| style="width:100%" border="0"

$$\displaystyle e^{rx}(r^2+2{\pi}r+{\pi}^2)=0$$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Since $$\displaystyle e^{rx}=0$$, the ODE can be expressed as follows:
 * {| style="width:100%" border="0"

$$\displaystyle r^2+2{\pi}r+{\pi}^2=0$$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

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

$$\displaystyle (r+{\pi})^2=0$$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


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

$$\displaystyle r=-\pi,-\pi$$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

The final general solution for this problem is:


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


 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Check by substitution:


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

$$\displaystyle {y}'=-\pi c_{1}e^{-\pi x}+c_{2}[e^{-\pi x}-\pi xe^{-\pi x}]$$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


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

$$\displaystyle {y}''= {\pi}^2c_{1}e^{-\pi x}+c_{2}[-\pi e^{-\pi x}-\pi e^{-\pi x}+{\pi}^2xe^{-\pi x}]$$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Substituting $$\displaystyle y, {y}', {y}''$$, we reach:
 * {| style="width:100%" border="0"

$$\displaystyle \pi^{2}c_{1}e^{-\pi x}+c_{2}[-2\pi e^{-\pi x}+{\pi}^2xe^{-\pi x}]+2\pi[-\pi c_{1}e^{-\pi x}+c_{2}(e^{-\pi x}-\pi xe^{-\pi x})]+{\pi}^2[c_{1}e^{-\pi x}+c_{2}xe^{-\pi x}] =0$$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


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

$$\displaystyle \pi^{2}c_{1}e^{-\pi x}+\pi^{2}c_{2}xe^{-\pi x} -2\pi c_{2}e^{-\pi x}-2\pi^{2}c_{1}e^{-\pi x} -2\pi^{2} c_{2}xe^{-\pi x}+2\pi c_{2}e^{-\pi x} +\pi^{2}c_{1}e^{-\pi x} +\pi^{2}c_{2}xe^{-\pi x}=0$$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

All terms cancel on the left side of the equation giving:


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

$$\displaystyle 0=0$$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Therefore the solution is correct.

Author
Solved and typed by - Egm4313.s12.team4.jones 18:38, 6 February 2012 (UTC) Reviewed By - Team 4

Edited by - Egm4313.s12.team4.jones 18:38, 6 February 2012 (UTC)

Solution (Problem 6)
Using the following:


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

$$\displaystyle {y}=e^{rx}$$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


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

$$\displaystyle {y}'=re^{rx}$$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


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

$$\displaystyle {y}''=r^2e^{rx}$$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Now plug in $$\displaystyle {y},{y}',{y}''$$ into the original equation:
 * {| style="width:100%" border="0"

$$\displaystyle 10r^2e^{rx}-32{r}e^{rx}+25.6e^{rx}=0$$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Now factor out $$\displaystyle e^{rx}$$:
 * {| style="width:100%" border="0"

$$\displaystyle e^{rx}(10r^2-32r+25.6)=0$$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Since $$\displaystyle e^{rx}=0$$, the ODE can be expressed as follows:
 * {| style="width:100%" border="0"

$$\displaystyle 10r^2-32r+25.6=0$$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

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

$$\displaystyle r=1.6$$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

The final general solution for this problem is:


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


 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Check by substitution:


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

$$\displaystyle {y}'=1.6c_{1}e^{1.6x}+c_{2}[e^{1.6x}-1.6xe^{1.6x}]$$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


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

$$\displaystyle {y}''= {1.6}^2c_{1}e^{1.6x}+c_{2}[1.6e^{1.6x}-1.6e^{1.6x}+{1.6}^2xe^{1.6x}]$$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Substituting $$\displaystyle y, {y}', {y}''$$, we reach:
 * {| style="width:100%" border="0"

$$\displaystyle 25.6c_{1}e^{1.6x}+10c_{2}[3.2e^{1.6x}+2.56xe^{1.6x}]-51.2c_{1}e^{1.6x}-32c_{2}(e^{1.6x}+1.6xe^{1.6x})]+25.6[c_{1}e^{1.6x}+c_{2}xe^{1.6x}] =0$$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


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

$$\displaystyle 25.6c_{1}e^{1.6x}+32c_{2}e^{1.6x}+25.6c_{2}xe^{1.6x}-51.2c_{1}e^{1.6x}-32c_{2}e^{1.6x}-51.2c_{2}xe^{1.6x}+25.6c_{1}e^{1.6x}+25.6c_{2}xe^{1.6x} =0$$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

All terms cancel on the left side of the equation giving:


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

$$\displaystyle 0=0$$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Therefore the solution is correct.

Author
Solved and typed by - Egm4313.s12.team4.jones 18:38, 6 February 2012 (UTC) Reviewed By - Team 4

Edited by - Egm4313.s12.team4.jones 18:38, 6 February 2012 (UTC)

= Problem 5: Homogeneous Linear ODEs with Constant Coefficients =

Problem Statement
Problems found in the textbook on page 59.

Solution (2.5)
The following table, taken from page 58 of the textbook , will help with problems 16 and 17.

16. Find an ODE given the basis.


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

$$\displaystyle e^{2.6x}, e^{-4.3x}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.5.1)
 * }

This falls under Case I with the two roots being:


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

$$\displaystyle \lambda_1=2.6$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.5.2)
 * }


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

$$\displaystyle \lambda_2=-4.3$$ An ODE of this form is desired:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.5.3)
 * }


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

$$\displaystyle y''+ay'+by=0$$ The characteristic equation is:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.5.4)
 * }


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

$$\displaystyle \lambda^2+a\lambda+b=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.5.5)
 * }

Plugging $$\lambda_1$$ into the characteristic equation(2.5.5):


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

$$\displaystyle (2.6)^2+a(2.6)+b=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.5.6)
 * }

Simplifying:


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

$$\displaystyle 6.76+2.6a+b=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.5.7)
 * }

Plugging $$\lambda_2$$ into the characteristic equation:
 * {| style="width:100%" border="0"

$$\displaystyle (-4.3)^2+a(-4.3)+b=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.5.8)
 * }

Simplifying:


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

$$\displaystyle 18.49-4.3a+b=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.5.9)
 * }

Subtracting equation 2.5.9 from equation 2.5.7:


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

$$\displaystyle -11.73+6.9a=0$$ Solving for a:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.5.10)
 * }


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

$$\displaystyle a=1.7$$ Plugging a into equation 2.5.7:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.5.11)
 * }


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

$$\displaystyle 6.76+2.6(1.7)+b=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.5.12)
 * }

Solving for b:


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

$$\displaystyle b=-11.18$$ Plugging a and b to complete the ODE:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.5.13)
 * }


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


 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.5.14)
 * }

17. Find an ODE given the basis.


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

$$\displaystyle e^{-\sqrt5x}, xe^{-\sqrt5x}$$ This falls under Case II with the root being:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.5.15)
 * }


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

$$\displaystyle \lambda=-\frac{1}{2}a=-\sqrt5$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.5.16)
 * }

An ODE of this form is desired:


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

$$\displaystyle y''+ay'+by=0$$ The characteristic equation is:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.5.17)
 * }


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

$$\displaystyle \lambda^2+a\lambda+b=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.5.18)
 * }

Multiplying both sides of the equation by -2 yields:


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

$$\displaystyle a=2\sqrt5$$ An ODE of this form is desired:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.5.19)
 * }


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

$$\displaystyle y''+ay'+by=0$$ The characteristic equation is:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.5.20)
 * }


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

$$\displaystyle \lambda^2+a\lambda+b=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.5.21)
 * }

Plugging λ and a into the characteristic equation:


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

$$\displaystyle (-\sqrt5)^2+(2\sqrt5)(-\sqrt5)+b=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.5.22)
 * }

Simplifying:


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

$$\displaystyle 5-10+b=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.5.23)
 * }

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

$$\displaystyle b=5$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.5.24)
 * }

Plugging a and b to complete the ODE:


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


 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.5.25)
 * }

Author
Solved and typed by - Egm4313.s12.team4.jkl 15:52, 7 February 2012 (UTC) Reviewed By - Team 4

Edited by - Egm4313.s12.team4.jkl 15:52, 7 February 2012 (UTC)

=Problem 6=

Problem Statement
Realize spring-dashpot-mass systems in series as shown in Fig. p.1-4 with the similar characteristic as in (3) p.5-5, but with double real root $$\lambda=-3$$, i.e., find the values for the parameters k, c, m.

Solution
Recall the equation of motion for the spring dashpot mass system :
 * $$m(y''_k + \frac{k}{c}y'_k)+ky_k=f(t)

\!$$ Dividing the entire equation by m:
 * $$y''_k + \frac{k}{cm}y'_k+ \frac{k}{m}y_k=f(t)

\!$$ The characteristic equation for the double root :$$\lambda=-3\!$$ is:
 * $$ (\lambda+3)^2 = \lambda^2+6\lambda+9=0

\!$$ The corresponding L2-ODE-CC is :
 * $$y''+6y'+9=0\!$$

Matching the coefficients:
 * $$y''\Rightarrow 1=1\!$$


 * $$y'\Rightarrow \frac{k}{cm}=6\!$$


 * $$y\Rightarrow \frac{k}{m}=9\!$$

After algebraic manipulation it is found that the following are the possible values for k, c, and m:
 * k=18\!




 * m=2\!

Author
Solved and typed by - Egm4313.s12.team4.Lorenzo 20:04, 6 February 2012 (UTC) Reviewed By -

Edited by - Egm4313.s12.team4.Lorenzo 20:33, 6 February 2012 (UTC)

=Problem 7=

Problem Statement
Develop the MacLaurin series (Taylor series at t=0) for :
 * $$e^t\!$$
 * $$\cos t\! $$
 * $$\sin t\!$$

Solution
Recalling Euler's Formula :
 * $$e^{i\omega x}=\cos\omega x+i\sin\omega x\!$$

Recall the Taylor Series for $$e^x$$ at $$x=0$$ (also called the MacLaurin series)
 * $$e^{x}=\sum_{n=0}^{\infty}\frac{x^n}{n!}\!$$

By replacing x with t, the Taylor series for $$e^t\!$$ can be found:

even powers:
 * $$i^{2k}=(i^2)^k=(-1)^k\!$$

odd powers:
 * $$i^{2k+1}=(i^2)^k i=(-1)^k i\!$$

If we let $$x=it$$:
 * $$e^{it}=\sum_{n=0}^{\infty}\frac{i^n t^n}{n!}=\sum_{k=0}^{\infty}\frac{i^{2k} t^{2k}}{(2k)!}+\sum_{k=0}^{\infty}\frac{i^{2k+1} t^{2k+1}}{(2k+1)!}\!$$

Using the two previous equations:
 * $$e^{it}=\sum_{k=0}^{\infty}\frac{(-1)^{k} t^{2k}}{(2k)!}+\sum_{k=0}^{\infty}\frac{(-1)^{k} t^{2k+1}}{(2k+1)!}\!$$


 * $$\Rightarrow e^{it}=\cos t+i \sin t\!$$

Therefore, the first part of the equation is equal to the Taylor series for cosine, and the second part is equal to the Taylor series for sine as follows:



This problem was solved by using methods learned from the class notes.

Author
Solved and typed by - Egm4313.s12.team4.Lorenzo 20:32, 6 February 2012 (UTC) Reviewed By -

Edited by - Egm4313.s12.team4.Lorenzo 20:33, 6 February 2012 (UTC)

=Problem 8: General Solution for ODE=

Problem Statement
From ADVANCED ENGINEERING MATHEMATICS, Erwin Kreyszig, 10th Ed, Page 59, Problems 8 and 15 2.8A.
 * {| style="width:100%" border="0"

$$\displaystyle y'' + y' + 3.25y = 0 $$ 2.8B.
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.8A.1)
 * }
 * {| style="width:100%" border="0"

$$\displaystyle y'' + 0.54y' + (0.0729 + \pi )y = 0 $$ Find a general solution to the equations and check the solution using substitution.
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.8B.1)
 * }

Solution 2.8A
Equation 2.8A.1 can be written as a characteristic equation:
 * {| style="width:100%" border="0"

$$\displaystyle \lambda ^2 + \lambda + 3.25 = 0 $$ which is of the form
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.8A.2)
 * }
 * {| style="width:100%" border="0"

$$\displaystyle \lambda ^2 + a \lambda + b = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.8A.3)
 * }

We now check the discriminant by determining the value of
 * {| style="width:100%" border="0"

$$\displaystyle a^2 - 4b $$ Since the discriminant,
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.8A.4)
 * }
 * {| style="width:100%" border="0"

$$\displaystyle a^2 - 4b < 0 $$ making the discriminant negative we have complex conjugate roots. In this case the complex roots have a characteristic equation of:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.8A.5)
 * }
 * {| style="width:100%" border="0"

$$\displaystyle \lambda = -\frac{1}{2}a\pm wi $$ Where we can determine $$\ w $$ using the equation
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.8A.6)
 * }
 * {| style="width:100%" border="0"

$$\displaystyle w^2 = b - \frac{1}{4}a^2 $$ leaving $$\ w $$ as:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.8A.7)
 * }
 * {| style="width:100%" border="0"

$$\displaystyle w = \sqrt{b - \frac{1}{4}a^2} $$ Real solution for complex roots can be written in the form:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.8A.8)
 * }
 * {| style="width:100%" border="0"

$$\displaystyle y_1 = e^{-\frac{ax}{2}}\cos wx$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.8A.9)
 * }

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

$$\displaystyle y_2 = e^{-\frac{ax}{2}}\sin wx$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.8A.10)
 * }

This will yield a general solution of:
 * {| style="width:100%" border="0"

$$\displaystyle y = e^{-\frac{ax}{2}}(A \cos wx + B \sin wx)$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.8A.11)
 * }

where A and B are arbitrary. Solving for $$\ w $$ we get: $$\ w = \sqrt{3} $$

Therefore solving for $$\ \lambda $$ we get:
 * {| style="width:100%" border="0"

$$\displaystyle \lambda = -\frac{1}{2} \pm \sqrt{3}i $$ The final general solution for this problem is:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.8A.12)
 * }
 * {| style="width:100%" border="0"


 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.8A.13)
 * }

Check by substitution:

In order to check the solution we must find $$\displaystyle y' $$ and $$\displaystyle y'' $$ :

Now that we have
 * {| style="width:100%" border="0"

$$\displaystyle y= e^{-\frac{x}{2}}(A \cos \sqrt{3} x + B \sin \sqrt{3} x) $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.8A.14)
 * }


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

$$\displaystyle y'= -\frac{1}{2} e^{-\frac{x}{2}} (A \cos \sqrt{3} x + B \sin \sqrt{3} x) + e^{-\frac{1}{2}} (- \sqrt{3} A \sin \sqrt{3} x + \sqrt{3} B \cos \sqrt{3} x) $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.8A.15)
 * }

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

$$\displaystyle y''= \frac{1}{4} e^{-\frac{x}{2}} (A \cos \sqrt{3} x + B \sin \sqrt{3} x) - \frac{1}{2} e^{-\frac{x}{2}} (-\sqrt{3} A \sin \sqrt{3} x + \sqrt{3} B \cos \sqrt{3} x) $$ $$\color{White}y''= \color{Black} -\frac{1}{2} e^{-\frac{x}{2}} (- \sqrt{3} A \sin \sqrt{3} x + \sqrt{3} B \cos \sqrt{3} x) + e^{-\frac{x}{2}} (-3A \cos \sqrt{3} x - 3B \sin \sqrt{3})  $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.8A.16)
 * }

we can substitute these equations into the original equation (2.8A.1), After substitution we see that
 * {| style="width:100%" border="0"

$$\displaystyle 0 = 0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.8A.17)
 * }

Therefore the solution is correct.

Author
Solved and typed by - Egm4313.s12.team4.patel 16:23, 7 February 2012 (UTC) Reviewed By - Team 4

Edited by - Egm4313.s12.team4.patel 16:23, 7 February 2012 (UTC)

Solution 2.8B
Equation 2.8B.1 can be written as a characteristic equation:
 * {| style="width:100%" border="0"

$$\displaystyle \lambda ^2 + 0.54 \lambda + (0.0729 + \pi ) = 0 $$ which is of the form
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.8B.2)
 * }
 * {| style="width:100%" border="0"

$$\displaystyle \lambda ^2 + a \lambda + b = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.8B.3)
 * }

We now check the discriminant by determining the value of
 * {| style="width:100%" border="0"

$$\displaystyle a^2 - 4b $$ Since the discriminant,
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.8B.4)
 * }
 * {| style="width:100%" border="0"

$$\displaystyle a^2 - 4b < 0 $$ making the discriminant negative we have complex conjugate roots. In this case the complex roots have a characteristic equation of:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.8B.5)
 * }
 * {| style="width:100%" border="0"

$$\displaystyle \lambda = -\frac{1}{2}a\pm wi $$ Where we can determine $$\ w $$ using the equation
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.8B.6)
 * }
 * {| style="width:100%" border="0"

$$\displaystyle w^2 = b - \frac{1}{4}a^2 $$ leaving $$\ w $$ as:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.8B.7)
 * }
 * {| style="width:100%" border="0"

$$\displaystyle w = \sqrt{b - \frac{1}{4}a^2} $$ Real solution for complex roots can be written in the form:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.8B.8)
 * }
 * {| style="width:100%" border="0"

$$\displaystyle y_1 = e^{-\frac{ax}{2}}\cos wx$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.8B.9)
 * }

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

$$\displaystyle y_2 = e^{-\frac{ax}{2}}\sin wx$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.8B.10)
 * }

This will yield a general solution of:
 * {| style="width:100%" border="0"

$$\displaystyle y = e^{-\frac{ax}{2}}(A \cos wx + B \sin wx)$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.8B.11)
 * }

where A and B are arbitrary. Solving for $$\ w $$ we get: $$\ w = \sqrt{\pi} $$

Therefore solving for $$\ \lambda $$ we get:
 * {| style="width:100%" border="0"

$$\displaystyle \lambda = -0.27 \pm \sqrt{\pi}i $$ The final general solution for this problem is:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.8B.12)
 * }
 * {| style="width:100%" border="0"


 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.8B.13)
 * }

Check by substitution:

In order to check the solution we must find $$\displaystyle y' $$ and $$\displaystyle y'' $$ :

Now that we have
 * {| style="width:100%" border="0"

$$\displaystyle y= e^{-0.27} (A \cos \sqrt{\pi} x + B \sin \sqrt{\pi} x) $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.8B.14)
 * }


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

$$\displaystyle y'= -0.27e^{-0.27} (A \cos \sqrt{\pi} x + B \sin \sqrt{\pi} x) + e^{-0.27} (- \sqrt{\pi} A \sin \sqrt{\pi} x + \sqrt{\pi} B \cos \sqrt{\pi} x) $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.8B.15)
 * }

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

$$\displaystyle y''= 0.0729e^{-0.27} (A \cos \sqrt{\pi} x + B \sin \sqrt{\pi} x) - 0.27e^{-0.27} (- \sqrt{\pi} A \sin \sqrt{\pi} x + \sqrt{\pi} B \cos \sqrt{\pi} x) $$ $$\color{White}y''= \color{Black} - 0.27e^{-0.27} (- \sqrt{\pi} A \sin \sqrt{\pi} x + \sqrt{\pi} B \cos \sqrt{\pi} x) + e^{-0.27} (- \pi A \cos \sqrt{\pi} x - \pi B \sin \sqrt{\pi} x) $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.8B.16)
 * }

we can substitute these equations into the original equation (2.8B.1), After substitution we see that
 * {| style="width:100%" border="0"

$$\displaystyle 0 = 0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.8B.17)
 * }

Therefore the solution is correct.

Author
Solved and typed by - Egm4313.s12.team4.patel 16:23, 7 February 2012 (UTC) Reviewed By - Team 4

Edited by - Egm4313.s12.team4.patel 16:23, 7 February 2012 (UTC)

=Problem 9=

Problem Statement
Find and plot the solution for the L2-ODE-CC corresponding to (2.9.1).

In another figure, superpose 3 figures: (a) this figure, (b) the solution to Problem 1, (c) the solution to Problem 6.


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

$$\displaystyle \lambda^2 + 4 \lambda + 13 = 0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.9.1)
 * }

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

$$\displaystyle y(0) = 1$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.9.2)
 * }


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

$$\displaystyle y'(0) = 0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.9.3)
 * }

No excitation:
 * {| style="width:100%" border="0"

$$\displaystyle r(x) = 0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.9.4)
 * }

Solution to R2.1:
 * {| style="width:100%" border="0"

$$\displaystyle y = \frac{5}{7} e^{-2 x} + \frac{2}{7} e^{5 x}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.9.a)
 * }

Solution to R2.6:
 * {| style="width:100%" border="0"

$$\displaystyle y = e^{-3 x} (c_1 + c_2 x)$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.9.b)
 * }

Solution
Equation (2.9.1) corresponds to the following differential equation:


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

$$\displaystyle y'' + 4 y' + 13 = 0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.9.5)
 * }

To find the solution, solve for $$\displaystyle \lambda$$ using the quadratic formula:
 * {| style="width:100%" border="0"

$$\displaystyle \lambda = \frac{-b \pm \sqrt{b^2 - 4 a c}}{2 a}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.9.6)
 * }


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$$\displaystyle \lambda = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot13}}{2 \cdot 1}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.9.7)
 * }


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$$\displaystyle \lambda = -2 \pm 3i$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.9.8)
 * }

This leads to a general solution of


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$$\displaystyle y = e^{-2 x} (c_1 \cos 3 x + c_2 \sin 3 x)$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.9.9)
 * }

Use initial condition (2.9.2) to find the particular solution,


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$$\displaystyle y(0) = 1 = e^0 (c_1 \cos 0 + c_2 \sin 0)$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.9.10)
 * }


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$$\displaystyle c_1 = 1$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.9.11)
 * }

Find the derivative to apply the second initial condition (2.9.3).


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$$\displaystyle y' = e^{-2 x} (c_2 \cos 3 x - c_1 \sin 3 x) - 2 e^{-2x} (c_1 \cos 3 x + c_2 \sin 3 x)$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.9.12)
 * }


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$$\displaystyle y'(0) = 0 = e^0 (c_2 \cos 0 - \sin 0) - 2 e^0 (\cos 0 + c_2 \sin 0)$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.9.13)
 * }


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$$\displaystyle 0 = c_2 - 2$$
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 * style="width:95%" |
 * <p style="text-align:right">(2.9.14)
 * }


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$$\displaystyle c_2 = 2$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.9.15)
 * }

The solution to the initial value problem:


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


 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.9.16)
 * }

Plot of the solution (2.9.16):
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 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.9.17)
 * }

Plot of the solution (2.9.16), plot of (2.9.a), and plot of (2.9.b).
 * {| style="width:100%" border="0"


 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.9.18)
 * }

Author
Solved and typed by - Egm4313.s12.team4.friedman 02:01, 8 February 2012 (UTC) Reviewed By - Team 4

Edited by - Egm4313.s12.team4.friedman 02:01, 8 February 2012 (UTC)

=Notes and References=