a) Show that $f^2 - (f')^2 = 0$
I tried to solve by doing $f=\pm f'$ not sure what to do from here.
Suppose $f$ satisfies $f '' - f = 0$ and $f (0) = f '(0) = 0.$ Prove that $f$ is the null function of the following mode:
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calculus
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1Not sure if the title and body are in good agreement – 2017-01-24
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0This question has appeared in one of my calculus exams – 2017-01-24
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0Compute the derivative of $f^2 - f'^2$. – 2017-01-24
2 Answers
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Consider $g=f^2-(f')^2$. One has
$$g'(x)=2f(x)f'(x)-2f''(x)f'(x)=2f'(x)\left(f(x)-f''(x)\right)=0$$
So $g$ is constant and by assumption $g(0)=0$ so $f^2-(f')^2=0$
This means $f'(x)=f(x)$ or $f'(x)=-f(x)$.
Solving these two ODE one gets $f(x)=f(0)e^x=0$ or $f(x)=f(0)e^{-x}=0$
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$(f-f')(f+f')=0$
$\implies f'=f $ or $f'=-f$.
Try solving the two IVP's:
$1$. $f'(x)-f(x)=0$ with $f(0)=f'(0)=0$.
$2$. $f'(x)+f(x)=0$ with $f(0)=f'(0)=0$.
Both have a unique solution which $f(x)=0$.