Solving the Fourth-Order Differential Equation: d4y/dt4 - λ4 y= 0

In summary, the conversation discusses solving a linear equation with constant coefficients. The characteristic equation is mentioned and its factors are listed. The characteristic values are identified as lambda, negative lambda, imaginary lambda, and negative imaginary lambda. The person requesting help admits to not understanding and asks for a clearer explanation. The expert asks for more information about the person's understanding of linear differential equations and clarifies that they did not solve x^2 + lambda^2. The expert also questions why the person is attempting to solve differential equations if they do not know how to solve a simple quadratic equation.
  • #1
teeoffpoint
3
0
how can i solve d4y/dt4 - λ4 y= 0
 
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  • #2
Since that is a linear equation with constant coefficients, you can immediately write down its "characteristic equation", [itex]x^4- \lambxa^4= 0[/itex]. That factors as [itex](x^2- \lambda^2)(x^2+ \lambda^2)= (x- \lambda)(x+ \lambda)(x- i\lambda)(x+ i\lambda)= 0[/itex] and so has characteristic values [itex]\lambda[/itex], [itex]-\lambda[/itex], [itex]i\lambda[/itex], and [itex]-i\lambda[/itex]. Do you know what to do with those?
 
  • #3
hmm... i don't really understand. can u show me a clearer working? wat you mean by x^4- \lambxa^4= 0? thanks
 
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  • #4
What kind of explanation would you understand? Do you know anything at all about linear differential equations? I need to know what you do understand before I can explain much more.
 
  • #5
oh how did you solve x^2 + λ^2?
 
  • #6
I didn't solve [itex]x^2+ \lamba^2[/itex]. That is not an equation. I did solve [itex]x^2+ \lambda^2= 0[/itex] by the obvious method: I subtracted [itex]\lambda^2[/itex] from both sides to get [itex]x^2= -\lambda^2[/itex] and then took the square root of both sides. But now, in addition to my previous questions, which you still haven't answered, why are you trying to do differential equations if you don't know how to solve a simple quadratic equation?
 

1. What is a fourth-order differential equation?

A fourth-order differential equation is an equation that involves the fourth derivative of a function. It can be written in the form d^4y/dt^4 + p(t)d^3y/dt^3 + q(t)d^2y/dt^2 + r(t)dy/dt + s(t)y = f(t), where p, q, r, and s are functions of t and f(t) is the forcing function.

2. Why is solving fourth-order differential equations important?

Fourth-order differential equations can model many physical phenomena, such as vibrations, heat transfer, and fluid dynamics. Solving them allows us to understand and predict these phenomena, which is crucial in many fields of science and engineering.

3. What is the characteristic equation for a fourth-order differential equation?

The characteristic equation for a fourth-order differential equation of the form d^4y/dt^4 + p(t)d^3y/dt^3 + q(t)d^2y/dt^2 + r(t)dy/dt + s(t)y = 0 is λ^4 + p(t)λ^3 + q(t)λ^2 + r(t)λ + s(t) = 0.

4. How is the general solution of a fourth-order differential equation determined?

The general solution of a fourth-order differential equation can be determined by finding the roots of the characteristic equation and using these roots to construct the general solution. The number of linearly independent solutions will depend on the number of distinct roots.

5. What is the role of initial conditions in solving fourth-order differential equations?

Just like any other differential equation, initial conditions are necessary to determine the specific solution to a fourth-order differential equation. These initial conditions, such as the value of the function and its derivatives at a certain point, help to narrow down the possible solutions and find the one that satisfies all the given conditions.

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