User talk:Egm4313.s12.team4.patel

= Report 1 = EGM4313 Report 1

R1.5A
Kreyszig 10th Edition. Page 59 Problem 4

Given
From ADVANCED ENGINEERING MATHMATICS Erwin Kreyszig 10th Edition. Page 59 Problem 4
 * {| style="width:100%" border="0"

$$\displaystyle y'' + 4y' + (\pi^2 +4)y = 0 $$ Find a general solution to the above equation.
 * style="width:95%" |
 * style="width:95%" |
 * (1.5a.1)
 * }

Solution
The equation above has a characteristic equation:
 * {| style="width:100%" border="0"

$$\displaystyle \lambda^2 + 4\lambda + (\pi^2 +4) = 0$$
 * style="width:95%" |
 * style="width:95%" |
 * (1.5a.2)
 * }

which is of the form:
 * {| style="width:100%" border="0"

$$\displaystyle \lambda^2 + a\lambda + b = 0$$
 * style="width:95%" |
 * style="width:95%" |
 * (1.5a.3)
 * }

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

$$\displaystyle a^2 - 4b < 0$$ making it negative.
 * style="width:95%" |
 * style="width:95%" |
 * (1.5a.4)
 * }

The roots of the characteristic equation are:
 * {| style="width:100%" border="0"

$$\displaystyle \lambda = -\frac{1}{2}a\pm wi $$
 * style="width:95%" |
 * style="width:95%" |
 * (1.5a.5)
 * }

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

$$\displaystyle w^2 = b - \frac{1}{4}a^2$$
 * style="width:95%" |
 * style="width:95%" |
 * (1.5a.6)
 * }

Therefore real solutions can be written as:
 * {| style="width:100%" border="0"

$$\displaystyle y_1 = e^{-\frac{ax}{2}}\cos wx$$
 * style="width:95%" |
 * style="width:95%" |
 * (1.5a.7)
 * }

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

$$\displaystyle y_2 = e^{-\frac{ax}{2}}\sin wx$$
 * style="width:95%" |
 * style="width:95%" |
 * (1.5a.8)
 * }

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%" |
 * (1.5a.9)
 * }

where A and B are arbitrary.

Solving for $$\displaystyle w$$, the roots become:
 * {| style="width:100%" border="0"

$$\displaystyle \lambda = -2 \pm \pi i $$
 * style="width:95%" |
 * style="width:95%" |
 * (1.5a.10)
 * }

The final general solution for this problem is:
 * {| style="width:100%" border="0"


 * style="width:95%" |
 * style="width:95%" |
 * (1.5a.11)
 * }

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

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

$$\displaystyle y= e^{-2x}(A \cos \pi x + B \sin \pi x)$$
 * style="width:95%" |
 * style="width:95%" |
 * (1.5a.12)
 * }


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

$$\displaystyle y'= -2e^{-2x}(A \cos \pi x + B \sin \pi x) + e^{-2x}(-A \pi \sin \pi x + B \pi \cos \pi x)$$
 * style="width:95%" |
 * style="width:95%" |
 * (1.5a.13)
 * }

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

$$\displaystyle y''= 4e^{-2x}(A \cos \pi x + B \sin \pi x) - 2e^{-2x}(-A \pi \sin \pi x + B \pi \cos \pi x)$$ $$\color{White}y''= \color{Black} - 2e^{-2x}(-A \pi \sin \pi x + B \pi \cos \pi x) + e^{-2x}(-A \pi ^2 \cos \pi x - B \pi ^2 \sin \pi x)$$
 * style="width:95%" |
 * style="width:95%" |
 * (1.5a.14)
 * }

we can substitute these equations into the original equation,
 * {| style="width:100%" border="0"

$$\displaystyle y'' + 4y' + (\pi^2 +4)y = 0$$
 * style="width:95%" |
 * style="width:95%" |
 * (1.5a.15)
 * }

After substitution we see that
 * {| style="width:100%" border="0"

$$\displaystyle 0 = 0$$
 * style="width:95%" |
 * style="width:95%" |
 * (1.5a.16)
 * }

Therefore the solution is correct.

= Report 2 = EGM4313 Report 2

R2.8A
From ADVANCED ENGINEERING MATHEMATICS, Erwin Kreyszig, 10th Ed, Page 59, Problem 8

Given

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

$$\displaystyle y'' + y' + 3.25y = 0 $$ Find a general solution to the equation and check the solution using substitution.
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.8A.1)
 * }

Solution
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.

R2.8B
From ADVANCED ENGINEERING MATHEMATICS, Erwin Kreyszig, 10th Ed, Page 59, Problem 15

Given

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

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

Solution
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.