# Linear equation (differential equations) problem

• aero_zeppelin
In summary, the conversation revolves around solving the differential equation dP/dt + 2tP = P + 4t - 2, with the main steps being obtaining the integrating factor and using it to integrate both sides of the equation. The final solution is given as P = [2 e^(t^2 - t) + C ] / e^(t^2 - t), with the interval of definition being (-inf, inf).
aero_zeppelin

## Homework Statement

Solve:

dP/dt + 2tP = P + 4t - 2

## The Attempt at a Solution

I've done a couple of these but I'm not sure how to start with this one...

First I have to put it in standard form right?

dP/dt + P(2t - 1) = 4t - 2

Then obtain the integrating factor by integrating (2t - 1)

Right so far. Now calculate$$e^{\int 2t-1 dt}$$for your integrating factor.

Ok! I get:

e^(t^2 - t)

Then we include it in the equation and integrate both sides?

d [e^(t^2 - t) P ] / dt = e^(t^2 - t) * (4t - 2)

The right hand side integral is giving me problems though...

aero_zeppelin said:
Ok! I get:

e^(t^2 - t)

Then we include it in the equation and integrate both sides?

d [e^(t^2 - t) P ] / dt = e^(t^2 - t) * (4t - 2)

The right hand side integral is giving me problems though...

Try a u substitution: ##u = t^2-t## as your first step.

Oh god, I can't believe I missed that one hehe I was overcomplicating it...

I end up with:

P = [2 e^(t^2 - t) + C ] / e^(t^2 - t)

Interval of definition would be (-inf, inf) correct?

aero_zeppelin said:
Oh god, I can't believe I missed that one hehe I was overcomplicating it...

I end up with:

P = [2 e^(t^2 - t) + C ] / e^(t^2 - t)

Interval of definition would be (-inf, inf) correct?

Yes, since the denominator is never zero.

Thanks a lot!

## 1. What is a linear equation in terms of differential equations?

A linear equation in terms of differential equations is an equation that involves a dependent variable and its derivatives in a linear fashion. This means that the dependent variable and its derivatives are raised to the first power and are not multiplied or divided by each other.

## 2. How do you solve a linear equation in terms of differential equations?

To solve a linear equation in terms of differential equations, you must first rearrange the equation so that all the terms with the dependent variable and its derivatives are on one side and all other terms are on the other side. Then, you can use techniques such as separation of variables and integrating factors to solve for the dependent variable.

## 3. What is the order of a linear differential equation?

The order of a linear differential equation is determined by the highest derivative present in the equation. For example, a first-order linear differential equation has a first derivative, while a second-order linear differential equation has a second derivative.

## 4. Can a linear differential equation have multiple solutions?

Yes, a linear differential equation can have multiple solutions. This is because when solving a linear differential equation, you must also include a constant of integration. This constant can take on different values, resulting in different solutions to the equation.

## 5. What is the difference between a homogeneous and non-homogeneous linear differential equation?

A homogeneous linear differential equation is one in which all the terms have the same degree of the dependent variable and its derivatives. This means that the equation can be solved by substituting a single variable. On the other hand, a non-homogeneous linear differential equation has terms with different degrees, making it more complex to solve and requiring additional techniques such as the method of undetermined coefficients.

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