Vtyjkyuk

$P(x)$ is a polynomial function,

$$P(x)-P'(x) = 2x^2-8x$$

Determine the $P(1)$

I've tried to solve the polynomial for $0$

$$P(0) = 2x^2-8x$$

Here we get

$$x = 4 , 0$$

However, I believe that I've gone wrong so far.

A locally nilpotent linear operator on a vector space $V$ (not necessarily finite-dimensional) over a field $mathbbK$ is a linear operator $N:Vto V$ such that, for each $vin V$, there exists a nonnegative integer $r_v$ depending on $v$ such that
$$N^r_v(v)=0_V,.$$ Here, $N^0:=textid_V$. For such a linear operator $N$, the linear operator $textid_V-N:Vto V$ is bijective. To show this, we exhibit the inverse of $textid_V-N$ as $i(N):=sumlimits_r=0^infty,N^r$. This linear operator $i(N):Vto V$ is well defined because $N$ is locally nilpotent. It is clear that
$$left(textid_V-Nright),i(N)=textid_V=i(N),left(textid_V-Nright),.$$

Now, the derivative operator $D$, $Df:=f'$, is a locally nilpotent operator on the vector space $mathbbK[X]$ of polynomials in variable $X$ over an arbitrary field $mathbbK$, namely, for each $f(x)inmathbbK[x]$,
$$D^deg(f)+1fequiv0text for fnotequiv0,,text and D^0fequiv 0text for fequiv0,.$$
By convention, $D^0$ denotes the identity operator $I$.

Now, we go back to the problem. Observe that $$big((I-D),Pbig)(X)=P(X)-P'(X)=2X^2-8X,.$$ By the paragraph above, $$beginalignP(X)&=(I-D)^-1,left(2X^2-8Xright)=sum_r=0^infty,D^rleft(2X^2-8Xright)\ &=(2X^2-8X)+(4X-8)+4=2X^2-4X-4,.endalign$$ That is, $P(1)=-6$.

Let $P(x)$ be any n degree polynomial
$$P(x) = a_0 + a_1 x + a_2x^2 + ... + a_nx^n$$

$$a_0+a_1x+a_2x^2 + ... a_nx^n - (a_1+2a_2x + ...na_nx^n-1 = 2x^2 - 8x$$

$$(a_0-a_1) + (a_1-2a_2)x + (a_2 - 3a_3)x^2 + .... = 2x^2 - 8x$$

Now the r.h.s and l.h.s are two polynomials that are equal for all values of $x$. Therefore they must be equivalent. You should be able to proceed now..

Cheers!

as mentioned in the comment by @mfl that the polynomial must have degree two because if we take P as degree n polynomial then P-P' also has degree n thus n=2 because the right side polynomial is of degree 2.

you can solve it by taking a degree two polynomial $P(x)=ax^2+bx+c$

I got $b=c$ and $a=2$ and $2a-b=8$

thus $$P(x)=2x^2-4x-4$$
thus $$P(1)=2-4-4=-6$$
$hspace20pt$

Here is method two if want to solve this by differential equations

this is of the form $$fracdydx+P(x)y=Q(x)$$

$$fracdydx-y=8x-2x^2$$

$$P(x)=-1 space and space Q(x)=8x-2x^2$$
in this type of differential equations, we multiply the equation by integrating factor

integrating factor $$R(x)=e^int P(x) dx$$

thus $$R(x)=e^int -1 dx=e^-x$$

multiply by $R(x)$
$$e^-xfracdydx-e^-xy=e^-x(8x-2x^2)$$
$$fracd(e^-xy)dx=e^-x(8x-2x^2)$$
$$e^-xy=int e^-x(8x-2x^2) dx $$

apply integration by part two times on the right side you will get

$$e^-xy= e^-x(2x^2-4x-4) + C$$

thus $$y= 2x^2-4x-4+ C *e^x$$
since P is a polynomial C must be zero