Given the following $f(x) - f'(x) = x^3 + 3x^2 + 3x +1$
Calculate $f(9) = ?$
I have tried to play with different number of derivatives. Also tried to solve it by equations. Maybe there is some geometric meaning that could shade the light ?
I feel it is no complex problem at all.
Thanks for your solution
You don't have enough information unless you assume $f(x)$ is a polynomial. Without that $f(x)=Ce^x+x^3+6x^2+15x+16$ satisfies the requirement for any $C$ and $f(9)$ can be any real number. This comes from the usual differential equation technique of adding the general solution to the homogeneous part to any particular solution. Others have found the particular solution. The general solution to $f(x)-f'(x)=0$ is $f(x)=Ce^x$
Multiply both sides by $e^{-x}$. Then the equation becomes $$e^{-x}f(x)-e^{-x}f'(x)=(x+1)^3e^{-x}\iff (-e^{-x}f(x))' =\Big(-(x^3+6x^2+15x+16)e^{-x}\Big)'$$ (for the last part you can use integration by parts on the integral $\displaystyle \int(x+1)^3e^{-x}dx$) and therefore $e^{-x}f(x) =(x^3+6x^2+15x+16)e^{-x}+c$, that is $f(x)=x^3+6x^2+15x+16+ce^x$. Set $x=9$ and you get $f(9)$, depending on a constant $c$ (if you were given a value of $f$ on a point $x_0$, you would be able to find $c$ by substitution).
\begin{equation} f(x) = ax^4 + bx^3 + cx^2 + dx + e\\f'(x) = 4ax^3 + 3bx^2 + 2cx + d\\f(x) - f'(x) = ax^4 + (b-4a)x^3 + (c-3b)x^2 + (d-2c)x + e - d = x^3 + 3x^2 + 3x +1\\a=0, b=1, c=6, d=15, e=16 \end{equation}
So, the function is $f(x)=x^3 + 6x^2 + 15x + 16$ and $f(9) = 1366$
It is most convenient to search $f$ among the third degree polynomials with the leading coefficients $1$. Let $f_0 (x) = x^3 + a x^2 + b x + c$. Then the equation becomes $$(f_0 - f'_0) (x) := x^3 + (a - 3) x^2 + (b - 2a) x + (c - b) = x^3 + 3 x^2 + 3 x + 1.$$ Hence, we have $a = 6$, $b = 15$ and $c = 16$. Then $$f_0 (x) = x^3 + 6 x^2 + 15 x + 16.$$ But this is only one from a class of solutions. Now, let $f = f_0 + g$, where $(g - g') (x) \equiv 0$. This gives us $g (x) = c \exp x$ for any constant $c$. Thus, we have $$f (x) = x^3 + 6 x^2 + 15 x + 16 + c \exp x.$$ Hence, $f (9) = 1366 + c e^9$.
$f(x)$ must be a 3rd degree polynomial. I think we can agree on that much.
If $f(x)=ax^3+bx^2+cx+d$, then $f'(x)=3ax^2+2bx+c$.
Hence, $f(x)-f'(x)=ax^3+(b-3a)x^2+(c-2b)x+(d-c)$.
By equating parts, we get:
$$a=1$$
$$b-3a=3$$
$$c-2b=3$$
$$d-c=1$$
From this, we can get $a=1,b=6,c=15,d=16$.
$$f(x)=x^3+6x^2+15x+16$$
$$f(9)=9^3+6(9)^2+15(9)+16$$
$$f(9)=729+486+135+16=1366$$
As Hagen von Eitzen notes, there is an arbitrary constant that may arise:
$$\frac d{dx}f(x)+ce^x=f'(x)+ce^x$$
Subtracting the two will cancel the invisible $ce^x$, but it may have an affect.
And so, the solution, as you may call it, may more correctly be along the lines of $f(9)=1366+ce^9$, where $c$ could be anything.
Let $f(x)=g(x)+\sum_{r=0}^na_r x^r$ where $g(x)$ is a non-polynomial function
$f(x)-f'(x)$ $=g(x)-g'(x)+a_0-a_1+x(a_1-2a_2)+x^2(a_2-3a_3)+x^3(a_3-4a_4)+\cdots+x^{n-1}(a_{n-1}-na_n)+x^n$
$\implies n=3$
$\implies f(x)-f'(x) =g(x)-g'(x)+a_0-a_1+x(a_1-2a_2)+x^2(a_2-3a_3)+x^3(a_3)$
Comparing the coefficients of $x^3, a_3=1$
Comparing the coefficients of $x^2, a_2=3a_3+3=6$
Comparing the coefficients of $x, a_1-2a_2=3\iff a_1=2a_2+3=15$
Comparing the constants $a_0=a_1+1=16$
$\implies g(x)=g'(x)\iff \int\dfrac{d\ g(x)}{g(x)}=\int\ dx\iff\ln g(x)=x+c$
$\implies (x)=e^{x+c}+16+15x+6x^2+x^3$
