3 comments

1. 

   March 27, 2019 at 4:06 pm

   Evangelos says:

   And I forgot to include the little command at the top. I’ll get this blogging thing… eventually. Here’s my post, hopefully with equations this time!

   So, a little change of substituting y = f(x) gave the perfect visual cue to figure it out. Let’s take it from the top

   The mystery function in question is such that its ordinate set over [0, x], when rotated about the x-axis (or in our case, the t-axis), sweeps out a solid of rotation with a volume equal to x^2 f(x). Written otherwise:

      

   If we take the derivative with respect to x of both sides, we get

      

   Rearranging terms gives us the derivative of the function in terms of x and the function f(x)

      

   Now, this is where the visual cue comes into play. Let y = f(x), the above equation becomes

      

   Now, if we recall from section 8.25, a first-order homogeneous differential equation f(x, y) is such that

      

   So, we can test our above equation y’ to see if it is homogeneous.

      

   As such, y’ = f(x, y) is a homogeneous first order equation. Now, let y = vx, and y’ = v’x + v, our equation becomes

      

   Rearranging terms and using Leibniz notation gives us the separable equation:

      

      

   Factoring the left side of the equation gives us:

      

   Which we can change into partial fractions

      

   Solving for A and B gives us

      

   So, our separable equation is now:

      

   Integrating both sides

      

   And taking the exponential of each side

      

      

   Substituting back v = y/x

      

   Rearranging terms gives us y as a function of x and L

      

   And finally, we can use our initial condition f(1) = 2/pi to solve for L and give y as a function of x

      

   This gives us

      

   Which matches the answer in the back of the book.

2. 

   March 27, 2019 at 3:44 pm

   Evangelos says:

   Eureka!

   So, a little change of substituting y = f(x) gave the perfect visual cue to figure it out. Let’s take it from the top

   The mystery function in question is such that its ordinate set over [0, x], when rotated about the x-axis (or in our case, the t-axis), sweeps out a solid of rotation with a volume equal to x^2 f(x). Written otherwise:

   \begin{align*}
   \pi \int_{0}^{x} [f(t)]^2 dt = x^2 f(x)
   \end{align*}

   If we take the derivative with respect to x of both sides, we get

   \begin{align*}
   \pi [f(x)]^2 = 2x f(x) + x^2 f'(x)
   \end{align*}

   Rearranging terms gives us the derivative of the function in terms of x and the function f(x)

   \begin{align*}
   f'(x) = \frac{\pi [f(x)]^2 – 2xf(x)}{x^2}
   \end{align*}

   Now, this is where the visual cue comes into play. Let y = f(x), the above equation becomes

   \begin{align*}
   y’ = \frac{\pi y^2 – 2xy}{x^2} \\
   = f(x, y)
   \end{align*}

   Now, if we recall from section 8.25, a first-order homogeneous differential equation f(x, y) is such that

   \begin{align*}
   f(tx, ty) = f(x, y)
   \end{align*}

   So, we can test our above equation to see if it is homogeneous. Let y’ = f(x, y)

   \begin{align*}
   f(tx, ty) = \frac{\pi (ty)^2 – 2(tx)(ty)}{(tx)^2} \\
   = \frac{t^2(\pi y^2 – 2xy){t^2 x^2}
   = \frac{t^2}{t^2} \frac{\pi y^2 – 2xy}{x^2}
   = \frac{\pi y^2 – 2xy}{x^2}
   = f(x, y)
   \end{align*}

   As such, y’ = f(x, y) is a homogeneous first order equation. Now, let y = vx, and y’ = v’x + v, our equation becomes

   \begin{align*}
   v’x + v = \frac{\pi (vx)^2 – 2x(vx)}{x^2} \\
   = \frac{x^2}{x^2} \pi v^2 – 2v \\
   = \pi v^2 – 2v
   \end{align*}

   Rearranging terms and using Leibniz notation gives us the separable equation:

   \begin{align*}
   \frac{dv}{dx}x = \pi v^2 – 3v
   \end{align*}

   \begin{align*}
   \frac{dv}{\pi v^2 – 3v} = \frac{dx}{x}
   \end{align*}

   Factoring the left side of the equation gives us:

   \begin{align*}
   \frac{dv}{v(\pi v – 3)}
   \end{align*}

   Which we can change into partial fractions

   \begin{align*}
   \frac{dv}{v(\pi v – 3)} = \frac{A}{v} + \frac{B}{\pi v – 3}
   \end{align*}

   Solving for A and B gives us

   \begin{align*}
   A = -\frac{1}{3}; \quad B = \frac{\pi}{3}
   \end{align*}

   So, our separable equation is now:
   \begin{align*}
   \frac{1}{3} (\frac{-1}{v} + \frac{\pi}{\pi v + 3}) dv = \frac{dx}{x}
   \end{align*}

   Integrating both sides
   \begin{align*}
   \frac{1}{3} log|\pi – \frac{3}{v}| = log|x| + C
   \end{align*}

   And taking the exponential of each side
   \begin{align*}
   (\pi – \frac{3}{v})^{1/3} = Kx \quad (where \quad K = e^C)
   \end{align*}

   \begin{align*}
   \pi – \frac{3}{v} = Lx^3 \quad (where \quad L = K^3 = e^{3C})
   \end{align*}

   Substituting back v = y/x
   \begin{align*}
   \pi – \frac{3x}{y} = Lx^3
   \end{align*}

   Rearranging terms gives us y as a function of x and L
   \begin{align*}
   y = \frac{3x}{\pi – Lx^3}
   \end{align*}

   And finally, we can use our initial condition f(1) = 2/pi to solve for L and give y as a function of x
   \begin{align*}
   \frac{2}{\pi} = \frac{3}{\pi – L} \\
   2\pi – 2L = 3\pi \\
   L = \frac{-\pi}{2} \\
   \end{align*}

   This gives us
   \begin{align*}
   y = \frac{3x}{\pi + \frac{\pi}{2} x^3} \\
   & = \frac{6}{pi} \frac{x}{2 + x^3}
   \end{align*}

   Which matches the answer in the back of the book.

3. 

   March 22, 2019 at 1:28 pm

   Evangelos says:

   This one has been driving me nuts… The setup is obvious, but getting from the initial equation to the function f(x) escapes me…

   The setup is as such

      

   And if we differentiate both sides with respect to x, we get

      

   Then, if we divide both sides by \begin \pi[f(x)] \end , we get

      

   And the rest, well, I haven’t been able to pull apart. Maybe soon…