Can someone find the solution of this differential equation:

$y'' + 4x = \sec 2x ?$

Any help would be highly appreciated.

  • $\begingroup$ Are you sure this isn't supposed to be $y''+4y=\sec 2x$? $\endgroup$ – Tartaglia's Stutter Mar 1 at 20:13
  • $\begingroup$ Yeah I am also supposing that it should be 4y instead of 4x but I have taken this question from some competitive previous examination so I am not sure about this. $\endgroup$ – Abdur Rehman Mar 1 at 20:15
  • $\begingroup$ Do you know the two constant variation method. $\endgroup$ – hamam_Abdallah Mar 1 at 20:33
  • 1
    $\begingroup$ If it is $x$ you just integrate twice in $x$. If it is $y$ you solve the homogeneous equation which gives $y_h(x)=A\cos 2x+B\sin 2x$. Then you find a particular solution using variation of parameters that is in the form $y_p(x)=A(x)\cos 2x+B(x)\sin 2x$. Your final solution is $y=y_h+y_p$. $\endgroup$ – GReyes Mar 1 at 20:33

$$y''=-4x+\sec(2x)$$ $$y'=-2x^2+\int \sec(2x)dx=-2x^2+\frac12\ln\Big(\tan(2x)+\sec(2x) \Big)+c_1$$ $$y=-\frac23 x^3+c_1x +\frac12\int \ln\Big(\tan(2x)+\sec(2x) \Big)dx +c_2$$ There is no closed form for the integral in terms of a finite number of elementary functions.

A closed form is complicated, involving special functions, namely the polylogarithm function. http://mathworld.wolfram.com/Polylogarithm.html

It is much simpler to consider the integral itself as the closed form, understood as a function defined by an integral. This is a kind of special function, but not presently in the list of standard special functions.


If this comes from a textbook exercise, obviously there is a typo in the equation $\quad y''+4x=\sec(2x)$.

The equation is certainly $\quad y''+4y=\sec(2x)\quad$. It is not difficult to solve it in terms of elementary functions.

Also solving $\quad y''+4y'=\sec(2x)\quad$ involves another special function.


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