# Solve for the solution of the differential equation

• Kanashii
In summary, the conversation involved solving a differential equation using the method of variation of parameters. The poster provided a formula for solving for the particular solution, but encountered a discrepancy in their answer. Through further discussion and guidance, it was determined that the poster was solving for the specific solution to the homogeneous part instead of the entire equation. They were able to successfully find the correct particular solution by guessing and checking.
Kanashii

## Homework Statement

Solve for the solution of the differential equation and use the method of variation of parameters.
x - x = (e^t) + t

## Homework Equations

[/B]
W= (y2y1)-(y2y1)
v1 = integral of ( g(t) (y1) ) / W
v2 = integral of ( g(t) (y2) ) / W

## The Attempt at a Solution

[/B]
yc= c1 e^t + c2 e^-t
yp = v1 e^t + v2 e^-t

W= -2
v1 = integral of ((e^-t)(e^t + t)) / -2 = (1/2) (t - te^-t - e^-t)
v2 = integral of ((e^2t)(e^t + t)) / -2 = (1/2) (1/2 e^2t + te^t - e^t)

v1y1 = 1/2 te^t - 1/2 t - 1/2
v2y2 = 1/4 e^t - 1/2t + 1/2

adding these, yp = -t + 1/2 te^t + 1/4 e^t
yc + yp = (c1 e^t + c2 e^-t ) + (-t + 1/2 te^t + 1/4 e^t)
but the answer is (c1 e^t + c2 e^-t ) + (-t + 1/2 te^t)

I got an extra term. Where did I go wrong?
Thank you!

I do not see your reasoning - it looks like you have attempted to apply a method of solution by rote instead of understanding it.

Guessing - yc is supposed to be the general solution to the homogeneous part of the DE. In which case, yp, as you have written it, is a specific solution to the homogeneous part - but it is supposed to be a particular solution to the entire DE.
However, yc is not a solution to the homogeneous part.
On top of all that... the DE does not contain the variable "y" that you keep using.

Last edited:
I got the formula
from http://tutorial.math.lamar.edu/Classes/DE/VariationofParameters.aspx to solve for the particular solution of the DE.
In this problem, g(t) = e^t + t, y1 = e^t and y2 = e^-t .
The Wronskian of these two functions:
| e^t e^-t |
| e^t -e^-t |

= (e^t)(-e^-t) - (e^t)(e^-t) = -1-1 = -2
Then plugging in the values to the equation, I got 1/2 (t - te^-t - e^-t) for v1 (or u1 in the formula) and -1/2 (1/2 e^2t + te^t - e^t) for v2 (u2 in the formula).
Multiplying these by y1 and y2 respectively, I got these:
v1y1 = 1/2 te^t - 1/2 t - 1/2
v2y2 = 1/4 e^t - 1/2t + 1/2

So from what you have said, I have solved for the specific solution to the homogeneous part, not the DE.
And you have also said that yc is not the solution for the homogeneous part. I do not understand.

m^2 - 1 = 0 (is this not to solve the roots?)
m= 1, - 1
leading to yc = c1 e^1t + c2 e^-1t

... you have also said that yc is not the solution for the homogeneous part. I do not understand.
Well, first of all, the DE is ##\ddot x + x = t+e^t## ... there is no ##y## in that equation.
But you mean: ##x_c = Ae^t + Be^{-t}## ... where A and B are arbitrary constants
... is the general solution to: ##\ddot x + x = 0## ... that about right?
OK then: prove it.

 excuse me that's a bit flip.
If you want to understand how x_c is not a solution to the homogeneous part, just substitute it into the homogeneous part and see what happens.

trial solution: x= e^mt
x = me^mt
x = m^2 (e^mt)

x - x = 0
m^2 (e^mt) - e^mt = 0 (characteristic equation)
dividing both sides by e^mt,
m^2 = 1
m = 1, -1

for distinct real roots,
xc = A (e^ (m1)t + B (e^ (m2)t) = A e^t + B e^-t

?

Oh I misread - I thought it said ##\ddot x + x = 0## but that should be a minus sign.
So the second derivative should be the same as the function. Well done.

##x_c = Ae^t + Be^{-t}## ... you proved your case: well done :)
Now you need a particular solution to ##\ddot x - x = t + e^t## ... any solution will do...

Notice that if you now write ##x_p = Ce^t + De^{-t}## you are just rewriting the original ##x_c##, just using different labels for the "arbitrary constants"?

Since any solution will do, don't sweat the whole wronskian thing - just guess:
You need x_p to end up with a second derivative that has a "t+"something in it. t^3+something will do that...
You could try ##x_p=at^3+be^t## ... is that close to being a solution to ##\ddot x - x = t+e^t##?
How does it need to change to make it a solution?

Kanashii
Any solution will do?
ooohhh
I have tried substituting what I got from my solution into the DE.

xp = 1/2 te^t - t - 1/4 e^t
xp = 1/2 (te^t + e^t) - 1 - 1/4 e^t
xp = 1/2 (te^t + 2e^t) - 1/4 e^t

1/2 (te^t + 2e^t) - 1/4 e^t - ( 1/2 te^t - t - 1/4 e^t) = (1/2 te^t + 3/4 e^t) - (1/2 te^t - t - 1/4 e^t) = t + (4/4) e^t = t + e^t which is equal to the right hand side of the given DE.

Did I understand it right?

Last edited:
That checks out. Note: you originally (post #1) had:
[xp=] -t + 1/2 te^t + 1/4 e^t
Now you have
xp = 1/2 te^t - t - 1/4 e^t
... actually, since any particular solution will work, there is no special reason that your answer will agree in detail to theirs.

You should probably check that ##-t+te^t## (theirs) also works as a particular solution ...
BTW: also see: https://www.physicsforums.com/help/latexhelp/ for help writing equations.

Kanashii
Thank you very much!

No worries :)

Kanashii

## 1. What is a differential equation?

A differential equation is an equation that relates a function or a set of functions to its derivatives. It is used to describe the relationship between a quantity and its rate of change over time.

## 2. Why is it important to solve for the solution of a differential equation?

Solving for the solution of a differential equation allows us to understand the behavior of a system over time. This is crucial in many fields of science, such as physics and engineering, where the dynamics of a system need to be predicted and controlled.

## 3. What methods can be used to solve a differential equation?

There are various methods for solving a differential equation, such as separation of variables, substitution, and using an integrating factor. The choice of method depends on the type of differential equation and its complexity.

## 4. Can all differential equations be solved analytically?

No, not all differential equations can be solved analytically. Some equations are too complex and require numerical methods to approximate the solution. However, there are certain types of differential equations that have known analytical solutions.

## 5. Are there any real-world applications of solving differential equations?

Yes, there are many real-world applications of solving differential equations. For example, they are used in predicting population growth, studying the spread of diseases, modeling climate change, and analyzing financial markets. They are also essential in designing and optimizing systems in engineering and physics.

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