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!

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

$\displaystyle n^2+5n+6$

is correct but,

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

So the solution should look like

$\displaystyle y(r)=\frac{c_1}{r^2}+\frac{c_2}{r^3}$

Re: Solving Differential Equation

How would you solve this problem?

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