initial-value prob: 2y" + 5y' +3y =0, y(0) =3, y'(0)= -

maeveoneill

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Sep 24, 2005
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The question is to solve the initial value problem of 2y" + 5y' +3y =0, y(0) =3, y'(0)= -4

I have come up wit the general solution is y=c1e^(-3x/2) + c2xe^(-x)
=> y' = -3/2c1e^(-3x/2) - c2e^(-x)

y(0) = c1 + c2 =3
y'(0) = -3/2c1 - c2 = -4

i am stuck when trying to find c1 in terms of c2.

i get -2/3 or -1/2.. but the back of my book has c1 as 2 and c2 as 1

can anyone tell me how to ge these values for c1 and c2.
thanks
 
Re: initial-value problems

maeveoneill said:
The question is to solve the initial value problem of 2y" + 5y' +3y =0, y(0) =3, y'(0)= -4

I have come up wit the general solution is y=c1e^(-3x/2) + c2xe^(-x)
=> y' = -3/2c1e^(-3x/2) - c2e^(-x)

y(0) = c1 + c2 =3
y'(0) = -3/2c1 - c2 = -4

i am stuck when trying to find c1 in terms of c2.

i get -2/3 or -1/2.. but the back of my book has c1 as 2 and c2 as 1

can anyone tell me how to ge these values for c1 and c2.
thanks

Plese show us how you got your values for c_1 and c_2 - in detail.
 
-3/2c1- c2 = -4 from y'(o)
-3/2c1= -4 +c2
c1= (-4 +c2)/ (-3/2)
c1= 8/3 -(3/2)c2

8/3 - 3/2c2 + c2 = 3
-3/2 c2 + c2 = 3-8/3
c2 (-3/2 + 1) = 1/3
c2= (1/3)/(-3/2 +1)
c2= -2/3
 
maeveoneill said:
-3/2c1- c2 = -4 from y'(o)
-3/2c1= -4 +c2
c1= (-4 +c2)/ (-3/2)
c1= 8/3 -(3/2)c2 <---- That should be 2/3

8/3 - 3/2c2 + c2 = 3
-3/2 c2 + c2 = 3-8/3
c2 (-3/2 + 1) = 1/3
c2= (1/3)/(-3/2 +1)
c2= -2/3

Why don't you start with much simpler substitution

c2 = 3 - c1
 
Re: initial-value prob: 2y" + 5y' +3y =0, y(0) =3, y'(0

maeveoneill said:
I have come up wit the general solution is y=c1e^(-3x/2) + c2xe^(-x)
=> y' = -3/2c1e^(-3x/2) - c2e^(-x)-----> derivation is wrong, instead of - c2e^(-x) there should be +c2e^(-x)-c2xe^(-x)
 
Re: initial-value prob: 2y" + 5y' +3y =0, y(0) =3, y'(0

Hello, maeveoneill!

It must be your algebra . . . everything up to that point is correct.


Solve the initial value problem: \(\displaystyle 2y''\,+\,5y'\,+\,3y \:=\:0,\;\;y(0)\,=\,3,\;\;y'(0)\,=\, -4\)

I have come up with: \(\displaystyle \:y\:=\:C_1e^{-\frac{3}{2}x}\,+\,C_2e^{-x}\;\;\Rightarrow\;\;y' \:= \:-\frac{3}{2}C_1e^{-\frac{3}{2}x}\,-\,C_2e^{-x}\;\;\) . . . Right!

\(\displaystyle \begin{array}{ccc}y(0) &\;=\; & C_1\,+\,C_2 & \;=\;& 3 \\
y'(0) & \;=\; & -\frac{3}{2}C_1\,-\,C_2 & \;=\; & -4\end{array}\;\;\) . . . Correct!
. .
And you can't solve this system of equations?

Add the two equations: \(\displaystyle \;-\frac{1}{2}C_1 \:=\:-1\;\;\Rightarrow\;\;C_1\:=\:2\) . . . then: \(\displaystyle \:C_2\:=\:1\)

 
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