# solutionhw0 - Solutions to Homework Set 0 Differential...

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• PresidentHackerCaribou10582
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Solutions to Homework Set 0 Differential calculus : The point of the exercise was to make sure that you know how to differentiate integrals with respect to their limits: d da integraldisplay b a f ( x ) dx = - f ( a ) , d db integraldisplay b a f ( x ) dx = f ( b ) , and in general d dt integraldisplay b ( t ) a ( t ) f ( x,t ) dx = f ( b ( t )) ∂b ∂t - f ( a ( t )) ∂a ∂t + integraldisplay b ( t ) a ( t ) ∂t f ( x,t ) dx. Once you have this under control, the first problem is plug and chug: From y ( x ) = sin ω ( x - L ) ω sin ωL integraldisplay x 0 f ( t ) sin ωtdt + sin ωx ω sin ωL integraldisplay L x f ( t ) sin ω ( t - L ) dt we get y prime ( x ) = cos ω ( x - L ) sin ωL integraldisplay x 0 f ( t ) sin ωtdt + cos ωx sin ωL integraldisplay L x f ( t ) sin ω ( t - L ) dt. The two terms arising from the derivative of the integration limits have cancelled against each other. When we differentiate again, the two pieces arising from differentiating the factors outside the integral assemble to give - ω 2 y ( x ). The two terms from differentiating the integrals are y primeprime ( x ) + ω 2 y ( x ) = f ( x ) cos ω ( x - L ) sin ωx - cos ωx sin ω ( x - L ) sin ωL . The addition formula for sin( A + B ), now shows that this is equal to f ( x ) as required. For the second problem we have F prime ( x ) = K (0) f ( x ) + integraldisplay x 0 x K ( x - y ) f ( y ) dy = K (0) f ( x ) - integraldisplay x 0 f ( y ) y K ( x - y ) dy = K (0) f ( x ) - integraldisplay x 0 y [ f ( y ) K ( x - y )] dy + integraldisplay x 0 f prime ( y ) K ( x - y ) dy = K (0) f ( x ) - K (0) f ( x ) + f (0) K ( x ) + integraldisplay x 0 f prime ( y ) K ( x - y ) dy = f (0) K ( x ) + integraldisplay x 0 K ( x - y ) f prime ( y ) dy. He was not quite right therefore—unless f (0) happens to be zero.

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