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Problem:Let $f$ be twice differential function such that $f''(x)= -f(x)$ and $f''(x)=g(x)$.

If $h(x)=(f(x))^2 + (g(x))^2$ and $h(5)=3$. Find $h(10)$

Solution:

let $f(x)= asinx$

$f''(x)= -asinx$

$h(x) = 2 a^2 sin^2x$

Now h(5) = 3

So $\frac {3}{sin^25} = 2a^2$

Hence $h(10) = \frac {3}{sin^25}* sin^210$

$ = 3*4*cos^2 5 $

$ = 1 (approximately)$

Is there any other method for solving this question ??

I mean other than taking example.

rst
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    You just wrote $;f''(x)=-f(x)=g(x);$ . Did you really mean this? – DonAntonio Jun 23 '13 at 12:41
  • @DonAntonio It's far more plausible that OP mean $f'(x) = g(x)$, considering the equation that comes right after... but that's my guess. – Patrick Da Silva Jun 23 '13 at 12:42
  • @DonAntonio Yeah, same is given in the question – rst Jun 23 '13 at 12:42
  • @rst : I hope you notice that $f''(x) = -f(x)$ implies that $f$ is infinitely differentiable! So assuming '$f$ is twice differentiable' is actually stronger in your context. :P – Patrick Da Silva Jun 23 '13 at 12:42
  • I think there's something fishy here: you say a(the) solution is $;f(x)=\sin x;$, but then $$h(x)=f(x)^2+(f''(x))^2\le 2$$ and it's impossible to get $,3,$ here...what's going on? – DonAntonio Jun 23 '13 at 12:42
  • you can solve this as a differential equation – AnonymouseCat Jun 23 '13 at 12:43
  • @DonAntonio Then we can try to solve this question by taking $f(x)=asinx$ where a is real number – rst Jun 23 '13 at 12:49
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    Following up on DonAntonio's comment, another good solution is $f(x) = 0$ for all $x$, in which case also $h$ vanishes. You need to provide more information for the problem to be well-defined. – fuglede Jun 23 '13 at 12:49
  • I agree, @rst, yet you wrote $,\sin x,$ is the solution... – DonAntonio Jun 23 '13 at 12:53
  • @rst: Even in that case, you will still end up with multiple solutions. For example, both $a \sin(x)$ and $b\cos(x)$ give rise to solutions with $h(5) = 3$ by choosing $a$ and $b$ appropriately, but in those cases, $h(10)$ will have two different values. – fuglede Jun 23 '13 at 12:54

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