Question

solve:
y''-6y'+9y =3x^2

Answer 1

are there boundary conditions? y(0) and y'(0)?

Answer 2

no

Answer 3

the general solution is the sum of the homogeneous solution + particular solution.

homogeneous: solution of y'' - 6y' +9y = 0

particular: solution to y'' - 6y' + 9y = 3x^2

I'll get to it very quickly but do you know how to find the solution to either one?

Answer 4

I can find the first solution where y'' - 6y' +9y = 0, but i am getting confused with the particular solution

Answer 5

for y'' - 6y' + 9y = f(x)
the "form" of the particular solution we choose depends on what f(x) is.

Since f(x) in this case is a 2nd order polynomial, the solution will have the form:\[y_p(x) = ax^2 + bx + c\]. Our goal is to determine a, b and c.

To find it we plug it into the equation: y''(x) - 6y'(x) + 9y(x) = 3x^2

Let's find these first:\[y'(x) = 2ax + b\]\[y''(x) = 2a\]
Now plug:\[2a - 6(2ax + b) + 9(ax^2 + bx + c) = 3x^2\]
Simplified to:\[2a - 12ax -6b + 9ax^2 +9bx + 9c = 3x^2\]

This line tells us that:\[9ax^2 = 3x^2\]\[(9b - 12a)x = 0x\]\[2a - 6b + 9c = 0\]

First line: 9a = 3 ---> a = 1/3
Second line: 9b - 12a = 0 ---> b = 4/9
Third Line: 2(1/3) - 6(4/9) + 9c = 0 ---> c = 2/9

So the particular solution is:\[y_p(x) = \frac{ 1 }{ 3 }x^2 + \frac{ 4 }{ 9 }x +\frac{ 2 }{ 9 }\]

General Solution:
\[y_h(x) + y_p(x) = c_1 e^{3t} + c_2 te^{3t} + \frac{ 1 }{ 3 }x^2 + \frac{ 4 }{ 9 }x +\frac{ 2 }{ 9 }\]

Answer 6

ok i see how that works now, just one tiny question on how you got :
y′(x)=2ax+b
y′′(x)=2a
- other than that this has been huge help with understanding 2nd order problems

Answer 7

y(x) = ax^2 + bx + c
y'(x) = 2ax + b
y''(x) = 2a


glad i could help ^_^

Answer 8

right, just taking derivative. i should have seen that. thanks again