# Thread: Differential eq problem and area problem

1. ## Differential eq problem and area problem

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

2. ## Thoughts

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.