Recent content by HACR

The answer has four whereas i came up with only 3 since the roots are 1,2, and 3 respectively. I think W=2e^{6x}

Homework Statement solve for y(x). y"'-6y"+11y'-6=e^{4x} Homework Equations Wronskian determinant. Method of variations. The Attempt at a SolutionSupposing that [u', v', w'] are the solutions, wronskian det=W is 10e^{6x} By use of x_k=\frac{det(M_{k})}{det(x)}, I got...

What are the arguments of the limit in this case?

What happened to partial differential equation?

It says the shortest path is the straight line; however, the brachistochrone problem proves that it is actually a curved line on which a stone could accelerate more. OK, brachistochrone problem is discussed. But why is on page 1163, the Euler Lagrangian equal to...

To find the potential function, set f_{x}=\frac{2x \vec i}{(x^2+y^2)^{1/2}}Then take the integral w.r.t. x. which then rewrite f(x,y)=...+g(y). Then take the derivative of this w.r.t. y and equate it to the second one. Then f(x,y), the potential function satisfies the vector field.

Homework Statement The isoperimetric problem is of the finding the object that has the largest area with the equal amount of perimeters; however, how does the integral constrained by the arc length get maximized? http://mathworld.wolfram.com/IsoperimetricProblem.html Homework Equations...

oops.

wow that's great.

It turned out to be \lim_{n->\infty}\frac{1}{n}(n+1)(n+2)(n+3)...(2n)=\frac{e}{4}

Right, I should've fixed the limits of integration from 0to 1 to 1 to 2 when i did it. Thanks for point it out. Although I think the 1/n power indicated in the problem is not necessary.

Homework Statement Prove that if f(x) is continuous for 0<f(x)<1, then lim_{n->\infty}\frac{1}{n}[(n+1)(n+2)(n+3)...(2n)]^{\frac{1}{n}}=\frac{4}{e}. Homework Equationsf(x)=log(1+x) The Attempt at a Solution We know that...

Is it sin int(t)*t+cos(t)-1? since D(sin int(x))=sinc(x)=sin(x)/x, u=sin int(x). If you look a int^udv=uv-int^v*du, du is on the RHS.

and should multiply each by b/3 to get the area since it's an integral.so \frac{b}{3}(e^{-1}+e^{-4}+e^{-9})=F(3)~F(b)

"e^{-µ(x)} * ∫e^{µ(x)} q(x)dx,µ(x) = ∫p(x)dx,µ(x) = px" You said µ(x)=px, but p is a function of x, so I believe it's something else. Also I think you meant not the first order since the original diff. equ is already of first order but of first degree.