Respuesta :
Hello there. To solve this question, we'll have to remember some properties about functions and derivatives.
We want to determine any possible functions h and g, where h cannot equal g, but knowing that h is the derivative of g and g is the derivative of h.
Mathematically, we have that
[tex]h^{\prime}(x)=g(x)\wedge h(x)=g^{\prime}(x)\text{ }[/tex]The way we can solve this is to consider that both h and g are twice differentiable, hence you differentiate both sides of the first expression to get:
[tex]h^{\prime}^{\prime}(x)=g^{\prime}(x)[/tex]But knowing g'(x) is h(x), we get
[tex]h^{\prime}^{\prime}(x)=h(x)[/tex]Subtract h(x) on both sides of the equation
[tex]h^{\prime}^{\prime}(x)-h(x)=0[/tex]This is a second order linear homogeneous differential equation.
To solve it, assume that
[tex]h(x)=e^{\lambda x}[/tex]For a fixed constant λ
Taking its derivative twice, we get
[tex]\begin{gathered} h^{\prime}(x)=\lambda\cdot e^{\lambda x} \\ \\ \Rightarrow h^{\prime}(x)=\lambda^2\cdot e^{\lambda x} \end{gathered}[/tex]Plugging this into our equation, we get
[tex]\lambda^2\cdot e^{\lambda x}-e^{\lambda x}=0[/tex]Factor the exponential term
[tex]e^{\lambda x}\cdot(\lambda^2-1)=0[/tex]Since we know that
[tex]e^{\lambda x}>0\text{ for any }\lambda\in\mathbb{C}[/tex]We get that
[tex]\lambda^2-1=0[/tex]It is called the characteristic equation for this differential equation.
Solving it gives us:
[tex]\lambda=1\text{ and }\lambda=-1[/tex]Therefore the solutions to this equation are
[tex]h(x)=e^x\text{ and }h(x)=e^{-x}[/tex]But since this is a linear equation, any finite combination of these solutions are solutions to it, therefore we say that
[tex]h(x)=c_1\cdot e^x+c_2\cdot e^{-x}[/tex]Is the general solution to this equation for any real constants c1 and c2.
Taking the derivative of this function, we'll see that
[tex]g(x)=h^{\prime}(x)=c_1\cdot e^x-c_2\cdot e^{-x}[/tex]And the only case in which g(x) = h(x), that is what we don't want to have, is when
[tex]\begin{gathered} g(x)=h(x) \\ \\ c_1\cdot e^x-c_2\cdot e^{-x}=c_1\cdot e^x+c_2\cdot e^{-x} \\ \\ c_2=-c_2=0 \end{gathered}[/tex]So it is possible to find infinitely many functions h and g of x satisfying this condition, as long c_2 is not equal to zero.
Take, for example, the solution:
[tex]\begin{gathered} h(x)=3e^x+4e^{-x} \\ \\ g(x)=3e^x-4e^{-x} \end{gathered}[/tex]They are not equal, but satisfies the properties:
[tex]\begin{gathered} h(x)=g^{\prime}(x) \\ g(x)=h^{\prime}(x) \end{gathered}[/tex]As wanted.
Another way of solving is: Add the conditions as follows
[tex]h(x)+g(x)=h^{\prime}(x)+g^{\prime}(x)[/tex]Now make the following substitution:
[tex]f(x)=h(x)+g(x)[/tex]Since the differential operator is linear, it is true that
[tex]f^{\prime}(x)=h^{\prime}(x)+g^{\prime}(x)[/tex]Hence we have that
[tex]f(x)=f^{\prime}(x)[/tex]This is a first order linear homogeneous differential equation, but it is also a separable equation, that we can solve as:
Divide both sides of the equation by f(x) (knowing it is not identically equal to zero)
[tex]\dfrac{f'(x)}{f(x)}=1[/tex]Integrate both sides with respect to x, such that you get
[tex]\int\dfrac{f'(x)}{f(x)}\,\mathrm{d}x=\int1\,\mathrm{d}x[/tex]The left hand side of the integral equation can be calculated knowing that
[tex]\int\dfrac{\mathrm{d}x}{x}=\ln|x|+C,C\in\mathbb{R}[/tex]Hence we get
[tex]\ln|f(x)|+C[/tex]For the right hand side, we have the integral of a power of x. Remember that
[tex]1=x^0[/tex]Therefore applying the power rule:
[tex]\int\,x^n\,\mathrm{d}x=\dfrac{x^{n+1}}{n+1}+C,\text{ }n\text{ not equal to }-1[/tex]We get that
[tex]\int1\,\mathrm{d}x=\int x^0\,\mathrm{d}x=\dfrac{x^{0+1}}{0+1}+C_1=x+C_1[/tex]Hence the equation gives us
[tex]\ln|f(x)|+C=x+C_1[/tex]Subtract C on both sides of the equation
[tex]\begin{gathered} \ln|f(x)|=x+C_1-C \\ \\ \ln|f(x)|=x+K_1 \end{gathered}[/tex]K_1 is another arbitrary constant.
Raise both sides of the equation as a power of e, as follows:
[tex]\begin{gathered} \exp(\ln|f(x)|)=\exp(x+K_1) \\ \\ |f(x)|=K\exp(x)=Ke^x \end{gathered}[/tex]Here we applied the property of powers:
[tex]e^{a+b}=e^a\cdot e^b[/tex]Since e^(K_1) is another constant, we called it K.
In this case, since it is a modular equation for f(x), we say that K must be a positive constant.
Returning to the definition, we see that
[tex]h(x)+g(x)=Ke^x[/tex]Making g(x) = h'(x), we have another equation
[tex]h(x)+h^{\prime}(x)=Ke^x[/tex]This time it is not a homogeneous equation, although it can be solved in a very similar way. The solution will be
[tex]Ae^x+Be^{-x}[/tex]Just as we found before, for constants A and B.
We perform the same argument to say that h(x) will only be equal to g(x) if B = 0, therefore it is possible to find functions of this form satisfying these properties.
