Differential eq problem and area problem

**mayth** · May 24th 2009, 04:14 AM

A:
State a differential equation that has the solution:
y=3sin(4x+v)

B:
decide the number k that the area beneath the curve y=ke^-x on the first square will be 2e areaunits

Would be great if you showed how to solve these problems, step by step. thanks

**Media_Man** · June 5th 2009, 04:46 PM

A. $y=3\sin(4x+v) \rightarrow y'=12\cos(4x+v) \rightarrow y''=-48\sin(4x+v)=-16y$ , so the differential equation $y''=-16y$ has the solution you seek, where v is a constant of integration.

B. Area= $\int_a^b ke^{-x}dx=-ke^{-x}|_a^b=-k(e^{-b}-e^{-a})$ . I am not sure what is meant by "on the first square" here. You are seeking a k for which the Area equals $2e$ , but I do not know what this phrase is implying for choosing a and b.

Thread: Differential eq problem and area problem

LinkBack

Thread Tools

Display

Differential eq problem and area problem

Thoughts

Similar Math Help Forum Discussions

Differential problem

Differential problem

Differential problem

differential log problem

Surface area problem (in fact, more of a parametrization problem...)

Search Tags