The discussion focuses on solving the first-order linear differential equation dy/dt = y(3t^2 - 1) with the initial condition y(1) = -2. The solution process involves integrating both sides, leading to ln|y| = t^3 - t + C, which simplifies to |y| = Ae^(t^3 - t). The challenge arises in determining how to satisfy the initial condition with a negative value for y, given that e^C is always positive. The interpretation of the equation as a product rather than a composite function is also debated.
PREREQUISITESStudents studying differential equations, educators teaching calculus, and anyone seeking to understand the nuances of initial value problems in mathematical modeling.
Your work might not be right, but I might be misreading the problem. Does y((3t^2)-1) represent the product of y and 3t2 - 1? It could also be interpreted as a composite function rather than a product.maximade said:Homework Statement
dy/dt=y((3t^2)-1), y(1)=-2
Homework Equations
Basic integrals
The Attempt at a Solution
integrate on both sides: dy/y=dt((3t^2)-1)
Assuming for the moment that the right side of your equation is y * (3t2 - 1), the line above should bemaximade said:========>ln(y)=(t^3)-t+c
The two lines above should bemaximade said:========>y=e^((t^3)-t+c)
========>y=e^((t^3)-t)e^(c)
maximade said:I am not sure if its some e rule that I forgot but how do you get -2 from y(1), since e^C is always positive?