# First order linear differential equations

## Homework Statement

dy/dt=y((3t^2)-1), y(1)=-2

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?

Last edited by a moderator:

Mark44
Mentor

## Homework Statement

dy/dt=y((3t^2)-1), y(1)=-2

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?