Twice differentiable function f(x) satisfying $f(x)+f''(x)=2f'(x)$ Consider a twice differentiable function f(x) satisfying $f(x)+f''(x)=2f'(x)$ where $f(0)=0,f(1)=e$.
Find the value of $f'(-1)$ and $f''(2)$
I used the following concept $f(x)=\alpha e^{ax}-\alpha $ as it satisfies $f(0)=0$ and $\alpha =\frac{e}{e^a-1}$, upto this level I am satisfied with my steps. Now using $f(x)+f''(x)=2f'(x)$
$\alpha e^{ax}-\alpha+\alpha a^2e^{ax}=2\alpha ae^{ax} $ which is equal to $ e^{ax}-1+a^2e^{ax}=2 ae^{ax} $
I am not able to use it in my formula
 A: Hint: this is a second-order linear differential equation with constant coefficients.  Start by finding the general solution.
A: We can re-write the equation as $y''(x)-2y'(x)+y(x)=0$ where $y=f(x)$ for convenience. This is a second-order ordinary differential equation with constant coefficients as Robert said.
Looking for a solution of the form $y=e^{mx}$ leads to $e^{mx}(m^{2}-2m+1)=0$ so the auxiliary (characteristic) equation is $m^{2}-2m+1=(m-1)^2=0$ which has repeated root $m=1$. Hence the general solution is $f(x)=(A+Bx)e^{x}$. Now we can use the boundary conditions to find the constants A and B.
A: hint
Let $$g(x)=f(x)-f'(x)$$
the equation becomes
$$g'(x)=g(x)$$
then
$$g(x)=e^{x}+C$$
thus
$$f'(x)=f(x)-e^{x}+C$$
so
$$f(x)=\lambda.e^x-xe^{x}-C$$
$$f'(x)=\lambda.e^x-(1+x)e^{x}$$
$$f''(x)=2f'(x)-f(x)$$
It is to you to find $\lambda$ and $ C$ such that
$$f(0)=0\;\; and \;\; f(1)=e$$
A: $$f(x)+f''(x)=2f'(x)$$
Is equivalent to:
$$(e^{-x}f(x))''=0$$
Integrate twice. And apply initial conditions you're given.
