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Solve the differential equation (x + y3) + 6xy2y′ = 0

Use a change of variables to convert the following differential equation into a linear differential equation, and then solve the equation:

    \[ (x+y^3) + 6xy^2 y' = 0. \]


Incomplete.

2 comments

    • Evangelos says:

      Yes indeed, Mr. Tom. Now, for those of you who don’t get the reference, earlier in this chapter, we were introduced to a form of the linear differential equation called the “Bernoulli Equation”, where we could transform a nonlinear first order equation into a linear first order equation for a new, unknown function. For further reading, see section 8.5 exercise #13, RoRi posted the solution here:

      http://www.stumblingrobot.com/2016/01/30/find-all-solutions-of-a-given-initial-value-problem/

      Now, onto our current exercise. We have the implicit first order equation

          \begin{align*} (x + y^{3}) + 6xy^{2}y' = 0 \end{align*}

      Which we can re-write as follows

          \begin{align*} \frac{(x + y^{3})}{6xy^{2}} + y' &= 0 \\ \frac{1}{6y^{2}} + \frac{y}{6x} + y' &= 0 \\ y' + \frac{y}{6x} &= - \frac{1}{6} y^{-2} \end{align*}

      And following from the proof in exercise 8.5 #13, we define a new function v such that:

          \begin{align*} v &= y^{3}\\ v' &= 3y^{2}y' \end{align*}

      The equation

          \begin{align*} y' + \frac{y}{6x} &= - \frac{1}{6} y^{-2} \end{align*}

      becomes

          \begin{align*} \frac{v'}{3v^{2/3}} + \frac{v^{1/3}}{6x} &= - \frac{1}{6} v^{-2/3} \\ v' + \frac{v}{2x} &= - \frac{1}{2} \\ v' + P(x)v &= Q(x) \end{align*}

      Which is a linear first order equation. From Theorem 8.3, the solution to the above equation, v = g(x), with g(a) = b is as follows:

          \begin{align*} g(x) &= e^{-A(x)} [b + \int_{a}^{x} Q(t) e^{A(t)} dt] \end{align*}

      With

          \begin{align*} P(x) &= \frac{1}{2x} \\ Q(x) &= -\frac{1}{2} \\ A(x) &= \int_{a}^{x} P(t) dt \\ &= \int_{a}^{x} \frac{1}{2t} dt \\ &= \log(\sqrt{\frac{x}{a}}) \end{align*}

      Giving us:

          \begin{align*} v &= (\frac{a}{x})^{1/2}[b - \frac{1}{2}\int_{a}^{x} (\frac{t}{a})^{1/2} dt] \\ &= (\frac{a}{x})^{1/2}[b - \frac{1}{3\sqrt{a}}(x^{3/2} - a^{3/2})] \\ &= Cx^{-1/2} - \frac{1}{3}x \quad (where \quad C = b\sqrt{a} + \frac{1}{3}a^{3/2}) \\ &= y^{3} \end{align*}

      For x>0, as presented in the back of the book.

      Now, for a solution for all x, we take the linear equation in v

          \begin{align*} v' + \frac{v}{2x} &= - \frac{1}{2} \end{align*}

      And we see that the equation can be satisfied by a first degree polynomial

          \begin{align*} v &= Ax \\ v' &= A \\ \end{align*}

          \begin{align*} v' + \frac{v}{2x} &= A + \frac{A}{2}\\ &= -\frac{1}{2}\\ A &= -\frac{1}{3}\\ \end{align*}

          \begin{align*} v &= -\frac{1}{3}x\\ &= y^{3} \end{align*}

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