The discussion focuses on integrating the fourth-order differential equation \( \frac{d^4u}{dx^4} + K \frac{d^2u}{dx^2} + 6 = 0 \) over the interval \( 0 < x < 1 \). Participants emphasize that this equation has constant coefficients and suggest using integration by parts. The integration process involves splitting the equation into three integrals and applying the fundamental theorem of calculus. The characteristic equation is also highlighted as a crucial step in solving the differential equation.
PREREQUISITESMathematicians, engineering students, and anyone involved in solving higher-order differential equations will benefit from this discussion.
This is NOT a first order equation: you can't just integrate both sides. rockfreak667 told you to do it as an equation with constant coefficients. What is its characteristic equation?shuttleman11 said:But I'm lost with assigning u, v, du, dv to this equation? Also can I split it into three idfferent integrals added together or must I assign the values and differentiate the hole equation by parts? Not for sure if I'm being clear?
And what is \int d^2y/dx^2 dx?rock.freak667 said:\frac{d^4u}{dx^4} + K \frac{d^2u}{dx^2} + 6=0
Now you can just integrate everything with respect to x to get
\int \frac{d^4u}{dx^4}dx + \int K \frac{d^2u}{dx^2}dx + \int 6dx= \int 0dx
Now \int \frac{dy}{dx}dx=y+c where c is a constant. Now just use this idea to work out your problem.