First order linear differential equations

[maximade]

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)
========>ln(y)=(t^3)-t+c
========>y=e^((t^3)-t+c)
========>y=e^((t^3)-t)e^(c)

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?

 

[Mark44]

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)

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.

========>ln(y)=(t^3)-t+c

Assuming for the moment that the right side of your equation is y * (3t² - 1), the line above should be
ln |y| = t³ - t + C.

========>y=e^((t^3)-t+c)
========>y=e^((t^3)-t)e^(c)

The two lines above should be
|y| = e^(t³ - t + C)
|y| = Ae^(t³ - t), where A = e^C

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?