Do you even integrate bro?
A positive, real valued, continuously differentiable function f satisfies the following equation, [ f ( x ) ] 2 = ∫ 0 x ( ( f ( t ) ) 2 + ( f ′ ( t ) ) 2 ) d t + e 2 .
Find f ( 1 ) .
Give your answer to 3 decimal places.
The answer is 7.389.
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1 solution
f ′ ( x ) = f ( x ) gives f ( x ) = k e x , for any real k . The expression f ( x ) = e x + c differs from this.
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Since f ( 0 ) = e f ( 0 ) = k e 0 = k = e which ultimately yields f ( x ) = e × e x = e x + 1
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Why can't f ( 0 ) = − e ?
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@Shourya Pandey – Shourya, the problem specifically states that f(x) is positive-valued, which makes its range strictly non-negative over its entire domain. That's why the boundary condition is restricted to just f(0) = e.
Hope that helps!
Since f(x) is differentiable for all real x, let us differentiate both sides of the above equation to yield:
2 f(x) f'(x) = f(x)^2 + f'(x)^2;
or f'(x)^2 - 2 f(x) f'(x) + f(x)^2 = 0;
or [f'(x) - f(x)]^2 = 0;
or f'(x) = f(x);
or f(x) = exp(x+c), f(0) = e (i).
Plugging in the boundary condition in (i) ultimately yields f(x) = exp(x+1), which can be checked for validity in the original integral equation. Hence, f(1) = e^2 = 7.389.