Let f:(-1,1) rightarrow R be a differentiable | vedclass

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(D) Given the differential equation $(f^{\prime}(x))^4 = 16(f(x))^2$,we have $(f^{\prime}(x))^2 = \pm 4f(x)$.Since $(f^{\prime}(x))^2 \geq 0$,we must

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more than $4$

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(D) Given the differential equation $(f^{\prime}(x))^4 = 16(f(x))^2$,we have $(f^{\prime}(x))^2 = \pm 4f(x)$.Since $(f^{\prime}(x))^2 \geq 0$,we must have $4f(x) \geq 0$,so $f(x) \geq 0$.Thus,$(f^{\prime}(x))^2 = 4f(x)$,which implies $f^{\prime}(x) = \pm 2\sqrt{f(x)}$.If $f(x) > 0$,then $\frac{f^{\prime}(x)}{\sqrt{f(x)}} = \pm 2$. Integrating both sides,we get $2\sqrt{f(x)} = \pm 2x + C$. Since $f(0)=0$,$C=0$,so $\sqrt{f(x)} = \pm x$,which gives $f(x) = x^2$.We can construct solutions by patching $f(x) = x^2$,$f(x) = -x^2$ (only if $f(x) \geq 0$ is not strictly required,but here $f(x) \geq 0$ is forced by the square),and $f(x) = 0$.Specifically,for any $a, b \in [0, 1)$,we can define $f(x)$ piecewise using $x^2$ and $0$ such that it remains differentiable at $x=0$. Since there are infinitely many such combinations,the number of such functions is more than $4$.

Let $f:(-1,1) \rightarrow R$ be a differentiable function satisfying $(f^{\prime}(x))^4 = 16(f(x))^2$ for all $x \in (-1,1)$ and $f(0)=0$. The number of such functions is:

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