# Second Order Non Homogeneous DE

• utility
Everything else is just plugging and chugging, I can do that.In summary, to solve the given second order differential equation, one must first find the solution to the homogeneous equation by using the annihilator method. This will yield the general solution, which can then be used to find the particular solution by plugging it into the original equation. The particular solution can then be added to the general solution to obtain the final solution to the given differential equation.

## Homework Statement

Find the solution to :y''+2y'+y=t

## Homework Equations

Suppose y(t)=B1t2+B2t+B3

And I believe, Y(t)=Yh+Yp. That is the solution is equal to the solution to the homogenous equation, plus the particular solution.

## The Attempt at a Solution

First Solve the homogeneous equation:
y(t)=B1t2+B2t+B3
y'(t)=2B1t+B2
y''(t)=2B1

Substitute this back into the homogenous problem(y''+2y'+y=0) gives:

2B1+2(2B1t+B2)+B1t2+B2t+B3=0

Rearrange:

B1(t2+4t+2) +B2(t+2)+B3=0

Solution is t=-2 or B1,B2,B3=0

Now I'm unsure of what to do?

As for the particular solution say Yp:

Yp=ct, where c is a constant so,
Y'p=c
Y''p=0

Substitute this into :y''+2y'+y=t
0+2c+ct=t

c=?

Thanks for any help.

You should try using the annihilator method.

I got Yp=At+B
Yp'=A
Yp"=0 (A and B are constants)

plugging them back into the ODE :

2[A] + [At+B] = t

rearranging:

(2A+B)[1] + A[t] = t

so

A=1

2A+B=0

getting A=1 and B= -2

so then Yp= t-2

add it to Yh and you should get Y(t)

oh, and to find Yh just set the ODE = 0 and find the roots

utility said:

## Homework Statement

Find the solution to :y''+2y'+y=t

## Homework Equations

Suppose y(t)=B1t2+B2t+B3

And I believe, Y(t)=Yh+Yp. That is the solution is equal to the solution to the homogenous equation, plus the particular solution.

## The Attempt at a Solution

First Solve the homogeneous equation:
y(t)=B1t2+B2t+B3
y'(t)=2B1t+B2
y''(t)=2B1

That is not the correct thing to try for the homogeneous solution. Try $y = e^{rt}$.

utility said:

## Homework Statement

Find the solution to :y''+2y'+y=t

## Homework Equations

Suppose y(t)=B1t2+B2t+B3

And I believe, Y(t)=Yh+Yp. That is the solution is equal to the solution to the homogenous equation, plus the particular solution.

## The Attempt at a Solution

First Solve the homogeneous equation:
y(t)=B1t2+B2t+B3
y'(t)=2B1t+B2
y''(t)=2B1
No, that is NOT how you solve for Yh.

Substitute this back into the homogenous problem(y''+2y'+y=0) gives:

2B1+2(2B1t+B2)+B1t2+B2t+B3=0

Rearrange:

B1(t2+4t+2) +B2(t+2)+B3=0

Solution is t=-2 or B1,B2,B3=0

Now I'm unsure of what to do?
That's because $B1x^2+ B2x+ B3$ is NOT a solution to the homogeneous equation. Use $y= e^rt$, as LCKurtz suggested, to get the "characteristic equation" to solve for r.

As for the particular solution say Yp:

Yp=ct, where c is a constant so,
Y'p=c
Y''p=0

Substitute this into :y''+2y'+y=t
0+2c+ct=t

c=?

Thanks for any help.
You cannot just let yp= ct. If the right hand side is a power of t, yp can be a polynomial nof degree at most that power- try yp= ct+ d.

Actually the annihilator method is correct. In fact I checked with my professor on this. The annihilator method takes out the factor of "guessing" a solution for the particular solution.

I even had my professor solve this and she got the same answer as I did, especially since we just went over this section in class.

Annihilator method is actually quite simple for this problem.

As for Yh, what I mean by set the ODE to 0 is:
y''+2y'+y=0

Let y''=r^2, y'=r, y=constant

ending up with:
r^2+2r+1=0

Factoring, you should end up with
(r+1)(r+1)=0

where r=-1,-1

Since this is a repeated root you should use:
Yh= c1*e^(rt) + c2*t*e^(rt)

Once you have Yh, then find Yp (particular solution) and add it to Yh for your general solution Y.

Just remember, since you DON'T have initial conditions

(for example: y'(0)=0, y(0)=1)

Your general solution is going to have constants c1 and c2 since there are no initial conditions given

Sorry for the late reply, yes I made a bad mistake with respect to the characteristic equation . I got it now though thanks!

Edit: I misunderstood the "hint" on the assignment. Every second order DE of this form, will have that solution given a single root.

## What is a Second Order Non Homogeneous Differential Equation?

A second order non homogeneous differential equation is an equation that contains a second derivative of a variable, and also has a non-zero function on one side of the equation. This means that the equation cannot be simplified to a homogeneous form.

## What is the difference between a Second Order Non Homogeneous DE and a Second Order Homogeneous DE?

The main difference between these two types of differential equations is that a second order homogeneous DE has a zero function on one side of the equation, meaning it can be simplified to a homogeneous form. In contrast, a second order non homogeneous DE has a non-zero function on one side, making it more complex to solve.

## What are some real-world applications of Second Order Non Homogeneous DEs?

Second order non homogeneous DEs can be used to model physical phenomena such as oscillations, vibrations, and heat transfer in various systems. They are also commonly used in engineering and science fields to predict the behavior of systems and make accurate projections.

## What are the general steps for solving a Second Order Non Homogeneous DE?

The general steps for solving a second order non homogeneous DE are: 1) Identify the type of equation (linear or nonlinear), 2) Determine the order of the equation, 3) Find the complementary function by solving the corresponding homogeneous DE, 4) Find the particular solution using the method of undetermined coefficients or variation of parameters, and 5) Combine the complementary function and particular solution to form the general solution.

## What are some common techniques for solving Second Order Non Homogeneous DEs?

Some common techniques for solving second order non homogeneous DEs include the method of undetermined coefficients, variation of parameters, Laplace transforms, and power series methods. The specific technique used will depend on the form of the equation and the given initial conditions.