Find a polynomial f(x) satisfying the equation
xf''(x) + 3f(x) = 4x^2.

Question

Find a polynomial f(x) satisfying the equation 
xf''(x) + 3f(x) = 4x^2.

anonymous · Answer

Since f''(x) and f'(x) has the maximum power (order) 2. Hence we can say the ploynomial is of 3rd order. Let the polynomial be 
$$f(x) = ax ^{3}+bx ^{2}+cx+d$$

This gives, $$f'(x) = 3ax ^{2}+2bx+c$$
And differentiating once more, $$f''(x) = 6ax+2b$$
Hence, $$xf''(x)+3f'(x) = 6ax^{2}+2bx+9ax ^{2}+6bx+3c = 12ax ^{2}+8bx+3c$$

From the differential equation
$$xf''(x)+3f'(x) = 12ax ^{2}+8bx+3c = 4x ^{2} $$
=> $$12a = 4, 8b=0, 3c = 0 => a = \frac{1}{3}, b=0, c=0 $$
Hence the polynomial is 
$$f(x) = \frac{1}{3}x ^{2}+d$$

Where, d is any constant or constant of integration

anonymous · Answer

im so lost, my homework said it was wrong, any thing else that could be the answer?

anonymous · Answer

Is it $xf''(x) + 3f(x) = 4x^2?$   or     $xf''(x) + 3f'(x) = 4x^2?$

anonymous · Answer

just f(x)

anonymous · Answer

Find a polynomial f(x) that satisfies 
xf ''(x) + 3f (x) = 4x^2.

anonymous · Answer

Oh I see.

anonymous · Answer

yeah this is calculus... and im lostulus lol

anonymous · Answer

I wasn't focusing. It's doable.

anonymous · Answer

i sure hope so

anonymous · Answer

The answer is $f(x)=\frac{4}{3}x^2-\frac{8}{9}x$.

anonymous · Answer

wow... however did you do it?

anonymous · Answer

Here how it's done. We're going to assume that the solution is a quadratic polynomial, since we have the sum of the function and its derivative adds up to $4x^2$. So let $f(x)=ax^2+bx+c \implies f'(x)=2ax+b \text{ and } f''(x)=2a.$
Plug all these into the given DE to get

$2ax+3ax^2+3bx+3c=4x^2$, or 
$3ax^2+(2a+3b)x+3c=4x^2$.

From the above equation the left hand side should have a coffienet of $4$ for $x^2$, and all other coffienents must be $0$. That is
$3a=4$,    $2a+3b=0$ and     $3c=0$.
You can easily solve the system above and find $a=\frac{4}{3}, b=-\frac{8}{9}$ and $c=0$.
Thus $f(x)=\frac{4}{3}x^2-\frac{8}{9}x$.

anonymous · Answer

Don't hisetate to ask if there's anything you don't get.

anonymous · Answer

okay so how did you get 2ax +3ax^2 +3bx + 3c = 4x^2

anonymous · Answer

thanks for youre help, but I have to go somewhere, you can still explain if you would like too, and i promise you ill come back and read it. thanks again!

anonymous · Answer

We are given that 
$$xf''(x) + 3f(x) = 4x^2$$      
But based on our assumption $f(x)=ax^2+bx+c$ and $f''(x)=2a$. Substitute for f(x) and f''(x) in the equation to get that.

anonymous · Answer

@AnwarA The f(x) is a cubic polynomial