# Solving Differential Equation

• Oct 16th 2012, 12:16 PM
nivek0078
Solving Differential Equation
I'm not getting the correct answer here and I'm sure i've got something screwed up.
given: r^2y''+6ry'+6y=0 y(2)=0, y'(2)=1

This is my workout:
r^2y''+6ry'+6y=0 y(2)=0, y'(2)=1

-try y=r^n
y'=nr^(n-1)
y''=n(n-1)r^(n-2)
- plugin
r^2(n(n-1))+6(nr^(n-1))+6(r^n)=0
n(n-1)+6n+6=0
(n^2)-n+6n+6=0
(n^2)+5n+6=0
-where
n=3, n=2
-general equation
y(r)=C1r^(3n) + C2r^(2n)
-take derivative of general equation which gives
y'(r)=3C1(ln(r)+1)r^(3r) + 2C2(ln(r)+1)r^(2r)

-plugin value y(2)=0 into general equ and solve for C1 and C2
0=C1(2)^(3*2) + C2(2)^(2*2)
C1=-C2/4 C2=-4C1
-Plugin value y'(2)=1 into the derivative
1=3C1(ln(2)+1)2^(3*2) + 2C2(ln(2)+1)2^(2*2)
C1=-(2C2-1)/192*(ln(2)+1) C2=(1/2)-96C1*(ln(2)+1)

So what do I do to finish out the problem? From the book the answer is y=(11/3e)r^2 - ((11e^(2))/3)r^-1
Thank you for your help in advance!
• Oct 16th 2012, 01:09 PM
TheEmptySet
Re: Solving Differential Equation
Quote:

Originally Posted by nivek0078
I'm not getting the correct answer here and I'm sure i've got something screwed up.
given: r^2y''+6ry'+6y=0 y(2)=0, y'(2)=1

This is my workout:
r^2y''+6ry'+6y=0 y(2)=0, y'(2)=1

-try y=r^n
y'=nr^(n-1)
y''=n(n-1)r^(n-2)
- plugin
r^2(n(n-1))+6(nr^(n-1))+6(r^n)=0
n(n-1)+6n+6=0
(n^2)-n+6n+6=0
(n^2)+5n+6=0
-where
n=3, n=2
-general equation
y(r)=C1r^(3n) + C2r^(2n)
-take derivative of general equation which gives
y'(r)=3C1(ln(r)+1)r^(3r) + 2C2(ln(r)+1)r^(2r)

-plugin value y(2)=0 into general equ and solve for C1 and C2
0=C1(2)^(3*2) + C2(2)^(2*2)
C1=-C2/4 C2=-4C1
-Plugin value y'(2)=1 into the derivative
1=3C1(ln(2)+1)2^(3*2) + 2C2(ln(2)+1)2^(2*2)
C1=-(2C2-1)/192*(ln(2)+1) C2=(1/2)-96C1*(ln(2)+1)

So what do I do to finish out the problem? From the book the answer is y=(11/3e)r^2 - ((11e^(2))/3)r^-1
Thank you for your help in advance!

Everything is good until right after this line

$n^2+5n+6$

is correct but,

$(n+2)(n+3)=0 \iff n=-2 \quad n=-3$

So the solution should look like

$y(r)=\frac{c_1}{r^2}+\frac{c_2}{r^3}$
• Oct 19th 2012, 07:12 AM
Jason76
Re: Solving Differential Equation
How would you solve this problem?

Theta squared d Theta = sin(t + 0.2) dt ?