User:EGM4313.s12.team6.davis/R2

R2.3
This problem consists of finding a general solution to two linear, second order differential equations with constant coefficients. The problems are taken from the tenth edition of Erwin Kreyszig's Advanced Engineering Mathematics and can be found as problems 3 and 4 of problem set 2.2 on page 59.

Given:
$$y''+6y'+8.96y=0$$

Find:
The homogeneous solution to the differential equation.

Solution:
Use the method of undetermined coefficients, also known as the method of trial solution, to solve the differential equation.

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

Therefore:


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

So,


 * $$y''+6y'+8.96y=0$$

can be written as


 * $$e^{\lambda x}(\lambda^2+a\lambda+b)=0$$  where   $$a=6$$   and   $$b=8.96$$

Since the exponential $$e^{\lambda x}$$ can never equal zero,


 * $$\lambda^2+a\lambda+b=0$$

This equation can be solved for lambda using the quadratic equation :


 * $$\lambda =\frac{-a\pm \sqrt[]{a^2-4b}}{2}$$

Substituting values:


 * $$\lambda =\frac{-(6)\pm \sqrt[]{(6)^2-4(8.96)}}{2}$$

Which evaluates to:


 * $$\lambda =-3\pm \frac{\sqrt[]{0.16}}{2}$$


 * or


 * $$\lambda =-3\pm 0.2$$

Therefore, we have:


 * $$\lambda_{1}=-2.8$$

and
 * $$\lambda_{2}=-3.2$$

So,
 * $$y_{1}=e^{-2.8x}$$

and
 * $$y_{2}=e^{-3.2x}$$

Superimposing the two solutions gives the homogeneous solution:


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

$$y=C_{1}e^{-2.8X}+C_{2}e^{-3.2x}$$
 * style="width:35%; padding:10px; border:2px solid #8888aa" |
 * style="width:35%; padding:10px; border:2px solid #8888aa" |
 * 
 * }
 * }

Given:
$$y''+6y'+(\pi^2+4)y=0$$

Find:
The homogeneous solution to the differential equation.

Solution:
Use the method of undetermined coefficients, also known as the method of trial solution, to solve the differential equation.

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

Therefore:


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

So,


 * $$y''+6y'+(\pi^2+4)y=0$$

can be written as


 * $$e^{\lambda x}(\lambda^2+a\lambda+b)=0$$  where   $$a=6$$   and   $$b=\pi^2+4$$

Since the exponential $$e^{\lambda x}$$ can never equal zero,


 * $$\lambda^2+a\lambda+b=0$$

This equation can be solved for lambda using the quadratic equation :


 * $$\lambda =\frac{-a\pm \sqrt[]{a^2-4b}}{2}$$

Substituting values:


 * $$\lambda =\frac{-(6)\pm \sqrt[]{(6)^2-4(\pi^2+4)}}{2}$$

Which evaluates to:


 * $$\lambda =-3\pm \frac{\sqrt[]{-19.478}}{2}$$


 * or


 * $$\lambda =-3\pm 2.207i$$

Since the discriminant is less than zero we let $$\omega=2.207$$ and the homogeneous solution will be of the form:


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

Substituting values we have the homogeneous solution:


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

$$y=e^(Acos(2.21 x)+Bsin(2.21 x))$$
 * style="width:45%; padding:10px; border:2px solid #8888aa" |
 * style="width:45%; padding:10px; border:2px solid #8888aa" |
 * 
 * }
 * }

R2.4
This problem consists of finding a general solution to two linear, second order differential equations with constant coefficients. The problems are taken from the tenth edition of Erwin Kreyszig's Advanced Engineering Mathematics and can be found as problems 5 and 6 of problem set 2.2 on page 59.

Given:
$$y''+2\pi y'+\pi^2y=0$$

Find:
The homogeneous solution to the differential equation.

Solution:
Use the method of undetermined coefficients, also known as the method of trial solution, to solve the differential equation.

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

Therefore:


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

So,


 * $$y''+2\pi y'+\pi^2y=0$$

can be written as


 * $$e^{\lambda x}(\lambda^2+a\lambda+b)=0$$  where   $$a=2\pi$$ and $$b=\pi^2$$

Since the exponential  $$e^{\lambda x}$$   can never equal zero,


 * $$\lambda^2+a\lambda+b=0$$

This equation can be solved for lambda using the quadratic equation :


 * $$\lambda =\frac{-a\pm \sqrt[]{a^2-4b}}{2}$$

Substituting values we can see that the discriminant is zero:


 * $$\lambda =\frac{-2\pi\pm \sqrt[]{4\pi^2-4\pi^2}}{2}$$

Therefore, only one solution can be found from this equation:


 * $$y_{1}=e^{-\pi x}$$

For the second solution reduction of order must be used:

Let:


 * $$y_{2}=uy_{1}$$

Then,


 * $$y_{2}'=u'y_{1}+y_{1}'u$$

and
 * $$y_{2}=uy_{1}+2u'y_{1}'+y_{1}''u$$

Substituting these into the original equation gives:


 * $$(uy_{1}+2u'y_{1}'+y_{1}u)+a(u'y_{1}+y_{1}'u)+ buy_{1}=0$$

Which simplifies to:
 * $$uy_{1}'+u'(2y_{1}'+ay_{1})+u(y_{1}+ay_{1}'+by_{1})=0$$

Since,


 * $$2y_{1}'=-ae^{\frac{-a}{2}x}=-ay_{1}$$

and
 * $$y_{1}''+ay_{1}'+by_{1}=0$$

What remains is:


 * $$u''y_{1}=0$$

or
 * $$u''=0$$

Therefore,


 * $$u=C_{1}x+C_{2}$$

Now we let  $$C_{1}=1$$   and   $$C_{2}=0$$   so that:


 * $$u=x$$

Now,


 * $$y_{2}=xy_{1}=xe^{-\pi x}$$

Using  $$y_{1}$$   and   $$y_{2}$$   we have the general solution:


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

$$y=(C_{1}+C_{2}x)e^{-\pi x}$$
 * style="width:25%; padding:10px; border:2px solid #8888aa" |
 * style="width:25%; padding:10px; border:2px solid #8888aa" |
 * 
 * }
 * }

Given:
$$10y''-32y'+25.6y=0$$

Find:
The homogeneous solution to the differential equation.

Solution:
First we put the equation into standard form by dividing through by 10 so that the equation becomes:


 * $$y''-3.2y'+2.56y=0$$

Now we can use the method of undetermined coefficients, also known as the method of trial solution, to solve the differential equation.

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

Therefore:


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

So,


 * $$y''-3.2y'+2.56y=0$$

can be written as


 * $$e^{\lambda x}(\lambda^2+a\lambda+b)=0$$  where   $$a=-3.2$$ and $$b=2.56$$

Since the exponential  $$e^{\lambda x}$$   can never equal zero,


 * $$\lambda^2+a\lambda+b=0$$

This equation can be solved for lambda using the quadratic equation :


 * $$\lambda =\frac{-a\pm \sqrt[]{a^2-4b}}{2}$$

Substituting values:


 * $$\lambda =\frac{-(-3.2)\pm \sqrt[]{(-3.2)^2-4(2.56)}}{2}$$

or
 * $$\lambda =\frac{3.2\pm \sqrt[]{0}}{2}$$

Which evaluates to:
 * $$\lambda = 1.6$$

Therefore, only one solution can be found from this equation:


 * $$y_{1}=e^{1.6x}$$

For the second solution reduction of order must be used:

Let:


 * $$y_{2}=uy_{1}$$

Then,


 * $$y_{2}'=u'y_{1}+y_{1}'u$$

and
 * $$y_{2}=uy_{1}+2u'y_{1}'+y_{1}''u$$

Substituting these into the original equation gives:


 * $$(uy_{1}+2u'y_{1}'+y_{1}u)+a(u'y_{1}+y_{1}'u)+ buy_{1}=0$$

Which simplifies to:
 * $$uy_{1}'+u'(2y_{1}'+ay_{1})+u(y_{1}+ay_{1}'+by_{1})=0$$

Since,


 * $$2y_{1}'=-ae^{\frac{-a}{2}x}=-ay_{1}$$

and
 * $$y_{1}''+ay_{1}'+by_{1}=0$$

What remains is:


 * $$u''y_{1}=0$$

or
 * $$u''=0$$

Therefore,


 * $$u=C_{1}x+C_{2}$$

Now we let  $$C_{1}=1$$   and   $$C_{2}=0$$   so that:


 * $$u=x$$

Now,


 * $$y_{2}=xy_{1}=xe^{1.6x}$$

Using  $$y_{1}$$   and   $$y_{2}$$   we have the general solution:


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

$$y=(C_{1}+C_{2}x)e^{1.6x}$$
 * style="width:25%; padding:10px; border:2px solid #8888aa" |
 * style="width:25%; padding:10px; border:2px solid #8888aa" |
 * 
 * }
 * }