General solution of equation

c4l3b

New member
Joined
Jul 17, 2009
Messages
37
I was wondering if some one could check my answer here...currently working through Engineering Mathematics book. I am imposing the conditions I have two different ways, the latter I do not understand perhaps some guidance will be helpful.

The question is ;

Find the general solution of the equation

\(\displaystyle y''+2y'+y=0\)

Assume that y(0) = 3 and y'(0) = 1 to find particular solution.

-----------------------------------------------------------------------------------


Auxiliary equation \(\displaystyle m^{2}+2m-1=1\)

Roots \(\displaystyle m= 1, \;\ m= -1\)

So the general solution is \(\displaystyle y(x)=C_{1}e^{x}+C_{2}e^{-x}\) or \(\displaystyle y(x)=Ae^{x}+Be^{-x}\)

Now to find the particular solution I impose the conditions

y(0) = 3 gives 3 = A + B
y'(0) = 1 gives 1 = 1A - 1B

A + B = 3
3A + 3B = 9
-------------------
-B = 12
B = -12
A - 12 = 3
A = 3
\(\displaystyle y_p(x)=3e^{x}- 12e^{-x}\) >>>>>>>>Final answer


On the other hand if \(\displaystyle 3=C_{1}e^{1(0)}+C_{2}e^{-1(0)}\Rightarrow 3=C_{1}+C_{2}\)

\(\displaystyle C_{2}=3-C_{1}\)

Then for y' \(\displaystyle 1=1C_{1}e^{1(0)}+(3 - C_{1})e^{-1(0)}\Rightarrow 0=C_{1}-1\)

Calculating with all this \(\displaystyle C_1, C_2\) stuff is somewhat confusing, any advice? is my final answer correct?
 
The general solution is \(\displaystyle y=(C_{1}x+C_{2})e^{-x}\)

\(\displaystyle y'=(C_{1}-C_{1}x-C_{2})e^{-x}\)

Now, use the IC for y:

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

\(\displaystyle C_{2}=3\)

Solve for C1 in y':

\(\displaystyle 1=(C_{1}-C_{1}(0)-3)e^{-0}\)

\(\displaystyle C_{1}=C_{2}+1\Rightarrow C_{1}=4\)

\(\displaystyle \boxed{y=(4x+3)e^{-x}}\)
 
So what was the characteristic equation, was it a iqual root (repeated) r-1^2?
 
\(\displaystyle m^{2}+2m+1=(m+1)^{2}=0\)

m=-1 repeated. Multiplicity 2

I noticed you have \(\displaystyle m^{2}+2m-1=1\). That gives roots \(\displaystyle m=-(\sqrt{3}+1), \;\ m=\sqrt{3}-1\)

Is that just a typo?.
 
But on the other hand how does the general solution justifies the auxliary equation, it looks different?
 
c4l3b said:
I was wondering if some one could check my answer here...currently working through Engineering Mathematics book. I am imposing the conditions I have two different ways, the latter I do not understand perhaps some guidance will be helpful.

The question is ;

Find the general solution of the equation

\(\displaystyle y''+2y'+y=0\)

Assume that y(0) = 3 and y'(0) = 1 to find particular solution.

You have a homogeneous ODE (no forcing function on the right hand side) - so your particular solution is in fact 0.

You'll use the initial (or boundary) conditions to find the total solution (y[sub:2c663ipa]total[/sub:2c663ipa] = y[sub:2c663ipa]homogeneous[/sub:2c663ipa] + y[sub:2c663ipa]particular[/sub:2c663ipa])


-----------------------------------------------------------------------------------


Auxiliary equation \(\displaystyle m^{2}+2m-1=1\)

Roots \(\displaystyle m= 1, \;\ m= -1\)

So the general solution is \(\displaystyle y(x)=C_{1}e^{x}+C_{2}e^{-x}\) or \(\displaystyle y(x)=Ae^{x}+Be^{-x}\)

Now to find the particular solution I impose the conditions

y(0) = 3 gives 3 = A + B
y'(0) = 1 gives 1 = 1A - 1B

A + B = 3
3A + 3B = 9
-------------------
-B = 12
B = -12
A - 12 = 3
A = 3
\(\displaystyle y_p(x)=3e^{x}- 12e^{-x}\) >>>>>>>>Final answer


On the other hand if \(\displaystyle 3=C_{1}e^{1(0)}+C_{2}e^{-1(0)}\Rightarrow 3=C_{1}+C_{2}\)

\(\displaystyle C_{2}=3-C_{1}\)

Then for y' \(\displaystyle 1=1C_{1}e^{1(0)}+(3 - C_{1})e^{-1(0)}\Rightarrow 0=C_{1}-1\)

Calculating with all this \(\displaystyle C_1, C_2\) stuff is somewhat confusing, any advice? is my final answer correct?
 
that whats I have done.

unless \(\displaystyle m_1=m_2= -1\) \(\displaystyle y_p(x)=3e^{x}- 12xe^{-x}\) can you the spot difference?


Actually \(\displaystyle y_p(x)=3e^{x}- 4xe^{-x}\) ?
 
c4l3b said:
that whats I have done.

unless \(\displaystyle m_1=m_2= -1\) \(\displaystyle y_p(x)=3e^{x}- 12xe^{-x}\) can you the spot difference?


Actually \(\displaystyle y_p(x)=3e^{x}- 4xe^{-x}\) ?

If you need practice with these kinds of questions, check out http://www.dr-mikes-maths.com/ode-examples.html

It's a web page that provides an infinite supply of examples of second order linear, constant coefficient inhomogeneous differential equations, along with how to solve them.
 
Top