Question

2nd order DE question

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Answer 1

I need to find its general solution first, I assume i can just factorise it to find the general solution

Answer 2

and then solve the IVP

Answer 3

As in get the factorise and then get the roots

Answer 4

use caracteristic equation

Answer 5

call d/dx=D
d^2/dx^2=D^2
you'll get a quadratic eqn in D
find its solutions
\[3D ^{2}+2D-1=0\]
whose solutions are D=-1 and D=1/3
now use D=dy/dx
to get solutions in terms of y and x

Answer 6

My general solution is y=c1*e^(-1/3x)+c2*(e^-1x)

Answer 7

y=c1*e^(-1/3x)+c2*e^(-1x)

Answer 8

yep its correct
now just substitute the boundary conditions to get c1 and c2

Answer 9

i think it's wrong should be:
y=c1*e^(1/3x)+c2*(e^-1x

Answer 10

just a careless mistake with the 1/3

Answer 11

Oops, should get two equations are subbing in both boundaries?

Answer 12

ya just differentiate the y equation once to sub in the y prime condition

Answer 13

I got c1+c2=1 and c1-3c3=9
I assume I can solve using simultaneous equations from here

Answer 14

sorry c1-3c2=9

Answer 15

I think I got it c1=-3 c2=4

Answer 16

ya as long as u differentiated and sub correctly u shld get the ans