Solve Eq: y''(x)+(2/x)y'(x)+(w^2)y(x)=0

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In summary, the equation y''(x)+(2/x)y'(x)+(w^2)y(x)=0 is a second-order linear homogeneous differential equation with y as a function of x and w as a constant. The steps to solve this equation include identifying coefficients, rewriting in standard form, finding roots, writing the general solution, and using initial/boundary conditions to find the specific solution. The constant w affects the behavior of the solution and can be solved both analytically and numerically. This type of equation has applications in physics, engineering, economics, and finance.
  • #1
mumaga
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Dear All,

I am having a problem finding the solution for this eq:

y''(x)+(2/x)y'(x)+(w^2)y(x)=0

I know it is quite elementary! any pointers will be much appreciated!

thanks
 
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no body wants to help even with a suggestion! :(
 
  • #3
Hi !

The change of function y(x) = f(x)/x will transform the ODE into a very simple linear ODE with constant coefficients.
 

FAQ: Solve Eq: y''(x)+(2/x)y'(x)+(w^2)y(x)=0

What is the meaning of the equation y''(x)+(2/x)y'(x)+(w^2)y(x)=0?

This equation is a second-order linear homogeneous differential equation, where y is a function of x and w is a constant. It can be solved to find the function y(x) that satisfies the equation.

What are the steps to solve this equation?

1. Identify the coefficients of y''(x), y'(x), and y(x). In this case, they are 1, 2/x, and w^2 respectively.
2. Rewrite the equation in the standard form y''(x)+p(x)y'(x)+q(x)y(x)=0, where p(x) and q(x) are the coefficients.
3. Find the roots of the characteristic equation r^2+p(x)r+q(x)=0. In this case, the roots are -w and -w^2.
4. Write the general solution as y(x)=c1e^(-wx)+c2e^(-w^2x), where c1 and c2 are constants.
5. Use initial conditions or boundary conditions to solve for the constants and find the specific solution.

What is the significance of the constant w in this equation?

The constant w determines the behavior of the solution. It can affect the frequency, amplitude, and stability of the solution. A change in w can result in different types of solutions, such as oscillatory or non-oscillatory.

Can this equation be solved analytically or numerically?

This equation can be solved both analytically and numerically. Analytical solutions involve finding a closed-form expression for the function y(x), while numerical solutions involve using algorithms to approximate the solution.

What are some real-life applications of this type of equation?

This type of equation is commonly used in physics and engineering to model various systems such as oscillating springs, electrical circuits, and vibrating structures. It can also be used in economics and finance to model population growth and stock prices.

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