First order linear differential equations

In summary, the problem involves finding the solution to the differential equation dy/dt = y(3t^2 - 1), with the initial condition y(1) = -2, using basic integrals. The solution involves integrating both sides of the equation, assuming that y((3t^2)-1) represents the product of y and 3t^2 - 1, and results in the equation ln|y| = t^3 - t + C. The final solution is |y| = Ae^(t^3 - t), where A = e^C. The question about getting -2 from y(1) may require further clarification on the problem.
  • #1
maximade
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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?
 
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  • #2
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)
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:
========>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.
maximade said:
========>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
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?
 

FAQ: First order linear differential equations

What is a first order linear differential equation?

A first order linear differential equation is a mathematical equation that relates a function and its derivative. It can be written in the form y' + p(x)y = q(x), where p(x) and q(x) are functions of x.

How do you solve a first order linear differential equation?

To solve a first order linear differential equation, you can use the method of integrating factors. This involves multiplying both sides of the equation by a suitable integrating factor, which is typically e∫p(x)dx. This will reduce the equation to the form (e∫p(x)dxy)' = q(x)e∫p(x)dx, which can then be integrated to find the solution for y.

What is the general solution to a first order linear differential equation?

The general solution to a first order linear differential equation is a family of functions that satisfy the equation. It can be found by integrating both sides of the equation and including an arbitrary constant C in the solution. The value of C can be determined by applying initial conditions or boundary conditions to the equation.

How do first order linear differential equations relate to real-world problems?

First order linear differential equations are commonly used in physics, engineering, and other fields to model real-world problems. They can be used to describe the rate of change of a physical quantity over time, such as the growth of a population, the decay of a radioactive substance, or the flow of a fluid through a pipe.

What are some common applications of first order linear differential equations?

Some common applications of first order linear differential equations include modeling population growth, chemical reactions, electrical circuits, and heat transfer. They are also used in economics to model supply and demand, in biology to model growth and decay of biological systems, and in finance to model interest and investment rates.

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