# 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

• Hint: the characteristic equation is $r^2-2r+1=0$. The solution is of the form $c_1e^{r_1x}+c_2e^{r_2x}$. Jun 15, 2020 at 13:43

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 $$f(0)=0\;\; and \;\; f(1)=e$$

• @LutzLehmann Yes, sorry, i made the correction. Thanks. Jun 15, 2020 at 14:31

Hint: this is a second-order linear differential equation with constant coefficients. Start by finding the general solution.

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.

$$f(x)+f''(x)=2f'(x)$$ Is equivalent to: $$(e^{-x}f(x))''=0$$ Integrate twice. And apply initial conditions you're given.