[calculus - Higher Derivatives] Find A Polynomial F X That Satisfies The Equation

[Calculus - Higher Derivatives] Find a polynomial f(x) that satisfies the equation..?

Hi! I'm not sure I understand your problem exactly, so just to be sure, I'm going to restate it here. Find a polynomial f(x) satisfying the equation xf''(x) + f(x) = x², where f''(x) is the second derivative of f.

Ok, so let's look at this problem in a very general way. The most general way to express a polynomial f(x) is f(x) = a₀ + a₁x¹ + a₂x² + ... + a_nx^n. We can easily compute f'(x) = a₁ + 2a₂x + ... + n*a_n*x^(n-1) and f''(x) = 2a₂ + 6a₃x + ... + n(n-1)a_n*x^(n-2). Let's plug this into our equation to see where we get,

xf''(x) + f(x) = x²
x(2a₂ + 6a₃x + ... + n(n-1)a_n*x^(n-2)) + (a₀ + a₁x¹ + a₂x² + ... + a_nx^n) = x²
(2a₂x + 6a₃x² + ... + n(n-1)a_n*x^(n-1)) + (a₀ + a₁x¹ + a₂x² + ... + a_nx^n) = x²

Now, I am going to collect the coefficient of each x^k term,

a₀ + (a₁ + 2a₂)x + (a₂ + 6a₃)x² + (a₃ + 12a₄)x³ + ... + (a_(n-1) + n(n-1))x^(n-1) + a_n*x^n = x²

Now, we equate coefficients on the LHS and RHS of the x^k terms (all equal 0 except x²),

x°: a₀ = 0
x¹: a₁ + 2a₂ = 0
x²: a₂ + 6a₃ = 1
x³: a₃ + 12a₄ = 0
...
x^(n-1): a_(n-1) + n(n-1)a_n = 0
x^n: a_n = 0

So, right off the bat, we have a₀ = 0 and a_n = 0. We want to systematically solve for the other coefficients. Let’s start with the second to last equation (for x^(n-1)), as this has a_n in it. We substitute this in to get a_(n-1) + n(n-1)(0) = 0, which implies a_(n-1) = 0. Furthermore, as long as the RHS of the equation equals zero, we can continue the same process, and all of the coefficients will continue to be zero (for example, for x³, we will have a₄ = 0, so a₃ + 12(0) = 0, which implies a₃ = 0).

Things get interesting when we get to the x² term, which does NOT equal 0 on the RHS. We will have a₃ = 0, as I showed above, so a₂ + 6(0) = 1, which implies a₂ = 1. Then, we move to the x equation to get a₁ + 2a₂ = 0, so a₁ + 2(1) = 0, so a₁ = -2. Now, we have solved for everything! Here are the results:

a₀ = 0
a₁ = -2
a₂ = 1
a₃ = 0
a₄ = 0
…
a_n = 0.

Thus, our final function is f(x) = -2x + x². Let’s check to see if this works. It should be pretty clear that f’(x) = -2 + 2x, so f’’(x) = 2. We want to see if it solves xf’’(x) + f(x) = x². So, we have x(2) + (-2x + x²) = 2x -2x + x² = x², as desired!

Hope this helps, and please pick for best answer!

Find a Polynomial f(x) satisfying the equation xf ′′ (x)+101f(x)=102x^2 .?

Hello,

x.f"(x) + 101.f(x) = 102.x²

We are told f is a polynomial.
The highest degree of the equality is 2.

Thus let us suppose:
f(x) = ax² + bx + c    (where a, b and c are real)

Then
f'(x) = 2ax + b
f"(x) = 2a

And:
x.f"(x) + 101.f(x) = 102.x²
2ax + 101ax² + 101bx + 101c = 102x²
(101a – 102)x² + (101b + 2a)x + 101c = 0

This leads to the system:
{ 101a – 102 = 0
{ 101b + 2a = 0
{ 101c = 0

whose solution is:
{ a = 102/101
{ b = -204/10201
{ c = 0

Thus f(x) = (102/101).x² – (204/10201).x

Check:
f"(x) = 204 / 101

Thus:
x.f"(x) + 101.f(x)
  = (204/101).x   +    101.[(102/101).x² – (204/10201).x]
  = (204/101).x + [(102.x² – (204/101).x]
  = 102.x²
QED.

Conclusively:
f(x) = (102/101).x² – (204/10201).x

Regards,
Dragon.Jade :-)

Find a Polynomial f(x) satisfying the equation xf"(x)+71f(x)=72x^2?

The equation can be satisfied by a quadratic polynomial
.. f(x) = ax^2 +bx +c
Substituting this into your equation gives
.. x*2a +71(ax^2 +bx +c) = 72x^2
.. (71a)x^2 +(2a +71b)x +71c = 72x^2

Equating coefficients, we get
.. 71a = 72 . . . . . . . coefficients of x^2
.. 2a +71b = 0 . . . . coefficients of x
.. 71c = 0 . . . . . . . . constant terms
So we know immediately
.. a = 72/71
.. c = 0
and
.. 2*72/71 +71b = 0
.. b = -144/71^2 = -144/5041

Your function is
.. f(x) = (72/71)x^2 -(144/5041)x

Taylor polynomial approximation?

i will do the 2nd, you can mimic the work in doing the 1st :
y ' ' + e^y = 0
y ' ' ' + e^y y ' = 0
y '''' + e^y [y ' ] ² + e^y y' ' = 0
y ^(5th) + e^y [ y ']^3 + e^y { 2 y ' y ''} + e^y [ y '] y'' + e^y { y ' ' '} = 0

T5(x|0) = 1 +[0] x + [-e] / 2 x² + [0] x^3 + [e²] / 24 x^4 + [0] x^5....the values for the derivatives are in the [..]s and are gotten from taking derivatives of the original equation.