Find the function satisfying given conditions I have a question which states that$$f'(x)=g(x)$$$$g'(x)=-f(x)$$$$f(x)-f'(x)=4$$find$$ f^2(19)+g^2(19)$$I tried it letting $$f(x)=a_0x^n+a_1x^{n-1}+a_2x^{n-2}\cdots +a_n$$then$$f''(x)=-f(x)=n(n-1)a_0x^{n-2}+(n-1)(n-2)a_1x^{n-3}+\cdots +a_{n-2}$$comparing the coefficients of $x^n,x^{n-1},x^{n-2}\cdots$ gives that $f(x)=0$ but it is not write as it does not satisfy $f(x)-f'(x)=4$
I discussed it with my friends and they said you cannot let $f(x)$ a polynomial function since it is not given in the question.
So I want two things.
Am I wrong really??
If no what should I do next?
If yes , please help me solving this problem
 A: $$\begin{cases}f'=0f+1.g \\g'=-1.f+0g\end{cases}\\\to \\\begin{bmatrix}f \\g \end{bmatrix}'=\begin{bmatrix}0 & 1 \\-1 & 0 \end{bmatrix}\begin{bmatrix}f \\g \end{bmatrix}$$ so $$|\begin{bmatrix}0-\lambda & 1 \\-1 & 0-\lambda \end{bmatrix}|=0 \to \lambda^2+1=0 \to \lambda=\pm i  $$now 
$$f(x)=ae^{ix}+be^{-ix}\\g(x)=ce^{ix}+de^{-ix}$$if you apply $f'=g,g'=-f $you will find $$i(ae^{ix}-be^{-ix})=ce^{ix}+de^{-ix}\\\to \\ia=c\\-bi=d $$now It is 
easy to show  $f^2+g^2=const$ because $$f'=g,g'=-f \to ff'+gg'=0 \to \frac12 f^2+\frac12 g^2=const \\\to \\f^2+g^2=const$$
$$f^2+g^2=(\underbrace{ae^{ix}+be^{-ix}}_{f(x)})^2+(\underbrace{iae^{ix}-ibe^{-ix}}_{g(x)})^2=a^2e^{2ix}+b^2e^{-2ix}+2ab +i^2a^2e^{2ix}+i^2b^2e^{-2ix}-2i^2ab=4ab$$
now apply
$$f-f'=4 \to ae^{ix}+be^{-ix}-(i(ae^{ix}-be^{-ix}))=4\\e^{ix}(a+c)+e^{-ix}(b-d)=4 $$ now find the value of $f^2+g^2=\\
$by simplifying $a=ic , b=-id $ you will see $f^2+g^2=4^2$
A: If f'= g and g'= -f, then f''= g'= -f so we have the differential equation f''+ f= 0.  The general solution to that is f(x)= A cos(x)+ B sin(x) and then g(x)= f'(x)= -A sin(x)+ B cos(x).  The condition that f(x)- f'(x)= 4 is  A cos(x)+ B sin(x)+ A sin(x)- B cos(x)= (A- B)cos(x)+ (A+ B) sin(x)= 4 for all x.  Taking x= 0, cos(0)= 1 and sin(0)= 0 so we have A- B= 4.  Taking $x= \pi/2$, $cos(\pi/2)= 0$ and $sin(\pi/2)= 1$ so we have A+ B= 4.  The solution to A- B= 4, A+ B= 4 is A= 4, B= 0 so we must have f(x)= 4 cos(x), g(x)= f'(x)= -4 sin(x).
$f^2(19)+ g^2(19)= 16 cos^2(19)+ 16 sin^2(19)= 16(cos^2(19)+ sin^2(19))= 16$.
