# Initial value nonhomogeneous DE

• jbord39
In summary, the conversation discusses the process of solving an ODE with initial conditions using factoring and an annihilator. The resulting solution is checked by substituting it back into the original equation and verifying that it satisfies the given initial conditions.
jbord39

## Homework Statement

(D^3 - D^2 + D - 1)y = e^x

and y(0) = 0, y'(0) = 1, y"(0) = 0

D = d/dx

## The Attempt at a Solution

Factoring the left side of the equation gives :
(D^2 + 1)(D - 1)y = e^x

Which has roots of +/- i and 1.
So y(c) = Ae^x + Bsin(x) + Ccos(x)

The annihilator for e^x is (D-1)
and y(p) = Dxe^x = e^x[Dx]
So y(p)' = D[xe^x + e^x] = e^x[D(x+1)]
y(p)" = D[xe^x + 2e^x] = e^x[D(x+2)]
y(p)"' = D[xe^x + 3e^x] = e^x[D(x+3)]

Plugging into y"' - y" + y' -y = e^x
and dividing by e^x yields:

2D = 1 ; or D = (1/2)

So now y = Ae^x + Bsin(x) + Ccos(x) + (1/2)xe^x

Plugging into initial conditions:
y(0) = A + C = 0 ; OR A = -C

Now y' = Ae^x + Bcos(x) - Csin(x) + (1/2)(xe^x + e^x)
and y'(0) = 1 = A + B + (1/2)(0 + 1)
or A + B = (1/2)

Now y" = Ae^x - Bsin(x) - Ccos(x) + (1/2)(xe^x + 2e^x)
and y"(0) = 0 = A - C + (1/2)(0 + 2)
or A - C + 1 = 0

Solving for A, B, and C yields:
A = (-1/2)
B = 1
C = (1/2)

The solution is therefore:

y = (-1/2)e^x + sin(x) + (1/2)cos(x) + (1/2)xe^x

Does this look correct?

Thanks a bunch

Anyone know if I'm in the right direction?

Bump.

It looks fine. You know you can check your own work by taking the y you get and substituting back into the original ODE? That is satisfies the boundary conditions isn't hard to check either?

Dick said:
It looks fine. You know you can check your own work by taking the y you get and substituting back into the original ODE? That is satisfies the boundary conditions isn't hard to check either?

Thanks, it makes sense but I did not think of that.

## 1. What is an initial value nonhomogeneous differential equation?

An initial value nonhomogeneous differential equation is a type of differential equation that involves both a function and its derivatives, as well as an initial condition. The initial condition is a specific value given for the function at a given point, which is typically used to solve the equation and find the function's general solution.

## 2. How is an initial value nonhomogeneous differential equation different from a homogeneous one?

An initial value nonhomogeneous differential equation differs from a homogeneous one in that the nonhomogeneous equation includes a function with a non-zero constant, while the homogeneous equation does not. This constant is often referred to as the forcing function or the nonhomogeneous term.

## 3. What is the general solution to an initial value nonhomogeneous differential equation?

The general solution to an initial value nonhomogeneous differential equation is a combination of the complementary function (a solution to the associated homogeneous equation) and the particular integral (a particular solution to the nonhomogeneous equation). The general solution is obtained by adding these two components together.

## 4. How do you solve an initial value nonhomogeneous differential equation?

To solve an initial value nonhomogeneous differential equation, you can follow these steps: first, solve the associated homogeneous equation to find the complementary function. Then, find a particular solution to the nonhomogeneous equation by using a method such as undetermined coefficients or variation of parameters. Finally, combine the complementary function and the particular solution to obtain the general solution, and use the initial condition to find the specific solution.

## 5. What are some real-life applications of initial value nonhomogeneous differential equations?

Initial value nonhomogeneous differential equations are commonly used in various fields of science and engineering, including physics, biology, and economics. They are particularly useful in modeling systems that involve external forces or influences, such as fluid dynamics, population growth, and electrical circuits.

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