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.
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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 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?