# Need help with solving an 2nd Order Linear ODE

• Turion
In summary, the student attempted to solve the homework equation but was not sure where to go from there. They solved the equation by guessing a particular solution and multiplying by the independent variable. If a term appears in your complementary solution and then in your particular solution, your particular solution needs to be changed a bit.
Turion

## Homework Statement

$$\frac { { d }^{ 2 }x }{ d{ t }^{ 2 } } +{ w }^{ 2 }x={ F }_{ 0 }sinwt\quad \quad \quad \quad x(0)=0\quad \quad x'(0)=0$$

## The Attempt at a Solution

$$\frac { { d }^{ 2 }x }{ d{ t }^{ 2 } } +{ w }^{ 2 }x={ F }_{ 0 }sinwt\quad \quad \quad \quad x(0)=0\quad \quad x'(0)=0\\ \\ \\ x_{ c }={ c }_{ 1 }coswt+{ c }_{ 2 }sinwt\\ \\ \\ { x }_{ p }=Atsinwt+Btcoswt\\ { x }_{ p }'=wAtcoswt-wBtsinwt\\ { x }_{ p }''=-{ w }^{ 2 }Atsinwt-{ w }^{ 2 }Btcoswt\\ \\ \\ Sub\quad { x }_{ p },\quad { { x }_{ p } }',\quad and\quad { { x }_{ p } }''\quad into\quad the\quad DE\\ \\ \\ 0={ F }_{ 0 }sinwt$$

I'm not sure where to proceed from here.

Turion said:

## Homework Statement

$$\frac { { d }^{ 2 }x }{ d{ t }^{ 2 } } +{ w }^{ 2 }x={ F }_{ 0 }sinwt\quad \quad \quad \quad x(0)=0\quad \quad x'(0)=0$$

If you have
$$x'' + ax' + bx = f(t)$$
and the obvious particular solution candidate $g(t)$ is a solution of $x'' + ax' + bx = 0$, then try the candidate $tg(t)$.

pasmith said:
If you have
$$x'' + ax' + bx = f(t)$$
and the obvious particular solution candidate $g(t)$ is a solution of $x'' + ax' + bx = 0$, then try the candidate $tg(t)$.

I don't follow. A particular solution applies to heterogeneous differential equations but you're using the term with a homogeneous differential equation?

Turion said:
I don't follow. A particular solution applies to heterogeneous differential equations but you're using the term with a homogeneous differential equation?

If a term appears in your complementary solution and then in your particular solution, your particular solution needs to be changed a bit. If the term is already in the complementary solution, how can it possibly be part of the particular solution?

That's why a useful trick is to multiply the guessed particular solution by the independent variable ( usually ##t## ), which will yield a completely different particular solution. If that does not get rid of all the duplicate terms between your complementary solution and your particular guess, multiply the particular guess by the independent variable a second time.

If you're working with a second order ODE, you will never need to multiply more than twice ( i.e the highest power term introduced through multiplying would be ##t^2## ). If you're working with an nth order ODE, you never need to multiply more than ##n## times.

For an easy example, try solving ##y'' - 3y' - 4y = e^{-t}##. Your complementary solution won't be too bad to solve, but try guessing the particular solution as you did before without multiplying through by the independent variable. You'll notice no combination of coefficients will satisfy your equation and your particular solution guess will be incorrect.

Last edited:
Zondrina said:
If a term appears in your complementary solution and then in your particular solution, your particular solution needs to be changed a bit. If the term is already in the complementary solution, how can it possibly be part of the particular solution?

That's why a useful trick is to multiply the guessed particular solution by the independent variable ( usually ##t## ), which will yield a completely different particular solution. If that does not get rid of all the duplicate terms between your complementary solution and your particular guess, multiply the particular guess by the independent variable a second time.

If you're working with a second order ODE, you will never need to multiply more than twice ( i.e the highest power term introduced through multiplying would be ##t^2## ). If you're working with an nth order ODE, you never need to multiply more than ##n## times.

I did do that though:

Turion said:
$${ x }_{ p }=Atsinwt+Btcoswt$$

or am I missing something?

Turion said:
I did do that though:

or am I missing something?

I seem to have missed the t's. You however missed them when differentiating:

$$(t \cos(wt))' = \cos(wt) - wt \sin(wt)$$

etc.

Thanks. Solved but was incredibly long.

## 1. What is a 2nd Order Linear ODE?

A 2nd Order Linear Ordinary Differential Equation (ODE) is a mathematical equation that involves a dependent variable, its derivatives, and independent variables. It is called "linear" because all the terms in the equation are either constants or multiples of the dependent variable and its derivatives.

## 2. How do you solve a 2nd Order Linear ODE?

To solve a 2nd Order Linear ODE, you need to first rearrange the equation into standard form, where the dependent variable and its derivatives are on one side and all other terms are on the other side. Then, you can use various methods such as the method of undetermined coefficients, variation of parameters, or Laplace transform to find the solution.

## 3. What are the initial conditions for a 2nd Order Linear ODE?

The initial conditions for a 2nd Order Linear ODE are the values of the dependent variable and its first derivative at a specific point, usually denoted as x0. These initial conditions are necessary to find the particular solution to the ODE.

## 4. What is the order of a 2nd Order Linear ODE?

The order of a 2nd Order Linear ODE is 2, as it involves the second derivative of the dependent variable. The order of an ODE is determined by the highest derivative present in the equation.

## 5. What are some real-life applications of 2nd Order Linear ODEs?

2nd Order Linear ODEs have various applications in science and engineering, such as modeling the motion of a spring-mass system, predicting the growth of a population, or analyzing the behavior of electrical circuits. They are also commonly used in fields such as physics, chemistry, and biology to describe natural phenomena.

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