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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

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

4 Answers 4

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

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  • $\begingroup$ @LutzLehmann Yes, sorry, i made the correction. Thanks. $\endgroup$ Jun 15, 2020 at 14:31
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Hint: this is a second-order linear differential equation with constant coefficients. Start by finding the general solution.

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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.

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

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