Question

Solve the given ODE.
1. (D^3+3D^2+3D+1)y=x^2
2.(D^4-2D^3-8D^2)y=6cosx
3.(D^3+1)y=9e^(x/2)

Answer 1

@hartnn

Answer 2

idk how to do dis sorry :(

Answer 3

1. \[ (D^3+3D^2+3D+1)y=x^2\]

Answer 4

to get the general solution, you need to find the homogenous solution, and the particular solution,

first find the homogenous solution, to the homogenous equation:
\[(D^3+3D^2+3D+1)y=0\]

Answer 5

can you teach me how to get homogenous solution? :D

Answer 6

if we suppose \(y=e^{mx}\),
\[D(y) = me^{mx}=my\]and\[D^2(y)=m^2e^{mx}=m^2y\]
etc.

Answer 7

so \[(D^3+3D^2+3D+1)y=0\]
will become
\[(m^3+3m^2+3m+1)y=0\]
and \(y=e^mx\) is not zero. so we can divide it out of the equation

we get
\[m^3+3m^2+3m+1=0\]

can you solve this for values of \(m\)?

Answer 8

if if was a quadratic, we could use the quadratic formula,

but alas, it is a cubic

Answer 9

can you factor it somehow?

Answer 10

\[(m-m_1)(m-m_2)(m-m_3)=0\]

Answer 11

(think of Pascal's triangle)

Answer 12

is that the final answer?

Answer 13

nope, we have to find the solutions for \(m\) (all of them),

then we will use them to get the homogenous solution to the differential equation.

Answer 14

you mean solution set? how?

Answer 15

[ the particular solution is more difficult to find ]

Answer 16

dont get mad. this is actually my boyfriend's homework. he is taking engineering course and i dont know how to solve this because im not into engineering. im so sorry. i just want to help him.

Answer 17

how to find the particular solution?

Answer 18

probably best to use operator D, method

Answer 19

So far on our quest to find the homogenous solution, we have;
\[m^3+3m^2+3m+1=0\]
Which is a cubic equation.

While cubics are usually horrid, this one is quite simple

Answer 20

Pascal's triangle looks something like this:



where terms below are made from the sum of the two number above

Answer 21

if we look at the \(\color{brown}{\textit{coefficients}}\) of the equation
\[(\color{brown}1)m^3+(\color{brown}3)m2+(\color{brown}3)m+(\color{brown}1)=0\]

Answer 22

what will i do with the coefficents? start factoring?

Answer 23

yes the \(m\) equation ends up factoring very nicely

Answer 24

(3m-1) • (1)

(m-3) • (m2)

Answer 25

A quadratic
\[(x+1)^2=(x+1)(x+1) \\=x(x+1)+1(x+1)\\=x^2+2x+1 \\=(\color{brown}1)x^2+(\color{brown}2)x+(\color{brown}1)\]
notice how the coefficients are a line in Pascal's triangle?