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Discuss properties of the solution of a differential equation

Consider the differential equation

    \[ y' = \frac{2y^2 + x}{3y^2 + 5}. \]

Let y = f(x) be a solution to the equation with the initial condition f(0)= 0. Without attempting to solve this equation explicitly answer the following questions.

  1. Since f(0)= 0 we also have f'(0) = 0. Does f have a relative maximum, relative minimum, or neither at 0?
  2. If x \geq 0 then f'(x) \geq 0 and if x \geq \frac{10}{3} then f'(x) \geq \frac{2}{3}. Find positive numbers a and b such that

        \[ f(x) > ax - b \qquad \text{for each } x \geq \frac{10}{3}. \]

  3. Prove that

        \[ \lim_{x \to +\infty} \frac{x}{y^2} = 0. \]

  4. Prove that

        \[ \lim_{x \to +\infty} \frac{y}{x} = A \]

    for some finite number A and find the value of A.



  1. Evangelos says:

    a.) From Theorem 4.9 on page 189, we can determine if a local extreme is a maximum, minimum, or neither. So, (assuming the initial conditions are met), if we take the second derivative of y, at x = 0, with y(0) = 0 we get

        \begin{align*} y'' &= \frac{(2y^{2} + x)'(3y^{2} + 5) - (3y^{2} + 5)'(2y^{2} + x)}{(3y^{2} + 5)^{2}} \\ &= \frac{(4yy' + 1)(3y^{2} + 5) - (6yy')(2y^{2} + x)}{(3y^{2} + 5)^{2}} \\ &= \frac{1}{5} \\ \end{align*}

    Since this value is greater than 0, from Theorem 4.9 we can conclude that this extreme is a local minimum.

    b.) We know that for x >= 10/3, f'(x) >= 2/3. And we know that for x>0, y>0. So, we can use the point-slope form of a line to exhibit a and b such that

        \begin{align*} f(x) > ax - b \end{align*}


        \begin{align*} x_{0} &= \frac{10}{3} \\ y_{0} &= 0 \\ m &= \frac{2}{3} \end{align*}


        \begin{align*} y - 0 > \frac{2}{3}(x - \frac{10}{3}) \end{align*}


        \begin{align*} a &= \frac{2}{3} \\ b &= \frac{20}{9} \end{align*}

    c.) We can use the above values for a and b to show that this is the case. Since we know that for all x >= 10/3

        \begin{align*} y > \frac{2}{3}(x - \frac{10}{3}) \end{align*}

    We also know that for all x >= 10/3

        \begin{align*} y^{2} > \frac{4}{9}(x^{2} - \frac{20}{3}x + \frac{100}{9}) \end{align*}

    Now, if we take the following limit

        \begin{align*} \lim_{x\to+\infty} \frac{x}{\frac{4}{9}(x^{2} - \frac{20}{3}x + \frac{100}{9})} \end{align*}

    We can see that the limit goes to 0

    But we know that

        \begin{align*} y^{2} > \frac{4}{9}(x^{2} - \frac{20}{3}x + \frac{100}{9}) \end{align*}

    So if the above limit goes to zero as x goes to infinity, then

        \begin{align*} \lim_{x\to+\infty} \frac{x}{y^{2}} \end{align*}

    Goes to zero as well.

    d.) We can once again use the results from part (b.). Since we know that for all x >= 10/3

        \begin{align*} y > \frac{2}{3}(x - \frac{10}{3}) \end{align*}

    We can use little-o notation from section 7.9 to show that that

        \begin{align*} \lim_{x\to+\infty} \frac{y}{x} &= \lim_{x\to+\infty} \frac{\frac{2}{3}(x - \frac{10}{3}) + o(1)}{x} \\ &= \lim_{x\to+\infty} \frac{2}{3} -\frac{20}{9x} + o(\frac{1}{x}) \\ &= \frac{2}{3} \end{align*}

    • Evangelos says:

      I think I can clarify part (c.) by using o-notation there as well.

      So, if we have the limit

          \begin{align*} \lim_{x\to+\infty} \frac{x}{\frac{4}{9}x^{2} - \frac{20}{3}x + \frac{100}{9}} \end{align*}

      We can use o-notation to write the above limit as

          \begin{align*} \lim_{x\to+\infty} \frac{x}{\frac{4}{9}x^{2} + \frac{20}{3}x + o(x)} \end{align*}

      And we can use Theorem 7.8 (c.) to simplify terms, giving us

          \begin{align*} \lim_{x\to+\infty} \frac{9/4}{x + 15 + o(1)}   \end{align*}

      As x goes to plus infinity, the limit goes to zero.

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