I am studying reduction of order from here:
http://tutorial.math.lamar.edu/Classes/DE/ReductionofOrder.aspx But this tutorial uses homogeneous examples to explain reduction of order. Now I am trying to apply reduction of order to this question:

$(D^2+1)y=\sec^3(x)$ use $y=v\sin(x)$

Please guide me to solve this problem. Thanks in advance.

  • $\begingroup$ I assume it is $y′′+y$ $\endgroup$ Dec 28, 2013 at 14:04
  • $\begingroup$ I am not sure either $\endgroup$ Dec 28, 2013 at 14:04
  • $\begingroup$ Yes this is one of my homework questions and it is written as above and it says solve it by reduction of order. $\endgroup$ Dec 28, 2013 at 14:08
  • $\begingroup$ It is correct, $\sin(x)$ is a homogeneous solution. So, letting $y = v \sin x$, find the second derivative using the product rule, substitute back into the DEQ and solve. Do you know how to find the second derivative using the product rule? $\endgroup$
    – Amzoti
    Dec 28, 2013 at 14:14

2 Answers 2



We have:

$$y = v \sin x$$

This gives us:

  • $y' = v \ cos x + v' \sin x$
  • $y'' = -v \sin x + v' \cos x + v' \ cos x + v'' \sin x$

Substituting into the ODE and simplifying, yields the new DEQ:

$$v'' \sin x + 2 v' \cos x = (\sec x)^3$$

Now, solve this using an Integrating Factor.


$v(x) = c_1 \cot x + c_2 + \dfrac{\tan x}{2}$

Of course, recall that the final solution is given by $y = v \sin x$.

Please fill in the details as this hint steps you through the entire problem.

  • $\begingroup$ I have substituted first and second derivatives of $y$ into equation: $(v''*sin(x) + 2*v'*cos(x) - v*sin(x))*v*sin(x)=sec(x)^3$ but I can not simplify it. Can you help abit more? $\endgroup$ Dec 28, 2013 at 17:34
  • $\begingroup$ Recall, you have $y'' + y$ from $D^2 + 1$. That is likely your issue since you do not have $y'$, which would show up as $D$. Clear? $\endgroup$
    – Amzoti
    Dec 28, 2013 at 18:58
  • $\begingroup$ @OlcayErtaş: Your DEQ written out is $y'' + y = (\sec x)^3$. Is that clear? The $D's$ are a shorthand notation. $\endgroup$
    – Amzoti
    Dec 28, 2013 at 19:04
  • $\begingroup$ This is how axactly the question is written: $(D^2 + 1)y = sec(x)^3$ $\endgroup$ Dec 28, 2013 at 19:10
  • $\begingroup$ Which it should be interpreted as: $(y''+ y)y=sec(x)^3$ wright? $\endgroup$ Dec 28, 2013 at 19:11

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$$\xi \equiv y' + {\rm i}y.\quad\xi' = y'' + {\rm i}y' = y'' + {\rm i}\left(\xi - {\rm i}y\right) = y'' + y + {\rm i}\xi.\quad$ Then, $y'' + y = \xi' - {\rm i}\xi$.

You will recover $y$ as $y = \Im\xi$. $$ \xi' - \ic\xi = \sec^{3}\pars{x}\quad\imp\quad\totald{\bracks{\expo{-\ic x}\xi}}{x} =\expo{-\ic x}\sec^{3}\pars{x} $$ $$ \expo{-\ic x}\xi = \int\expo{-\ic x}\sec^{3}\pars{x} + \mbox{a constant}. $$ Let's say, $\underline{\tt\mbox{for example}}$, we know ${\rm y}\pars{0}$ and ${\rm y}'\pars{0}$. Then, we'll know $\xi\pars{0} = {\rm y}'\pars{0} + \ic{\rm y}\pars{0}$: \begin{align} &\expo{-\ic x}\xi\pars{x} - \xi\pars{0} = \int_{0}^{x}\expo{-\ic t}\sec^{3}\pars{t}\,\dd t \\[3mm]\imp\ {\rm y}\pars{x} &= {\rm y}\pars{0}\cos\pars{x} + {\rm y}'\pars{0}\sin\pars{x} + \Im\int_{0}^{x}\expo{\ic\pars{x - t}}\sec^{3}\pars{t}\,\dd t \\[3mm]&= {\rm y}\pars{0}\cos\pars{x} + {\rm y}'\pars{0}\sin\pars{x} + \int_{0}^{x}\bracks{% \sin\pars{x}\sec^{2}\pars{t} - \cos\pars{x}\,{\sin\pars{t} \over \cos^{3}\pars{t}}}\,\dd t \\[3mm]&= {\rm y}\pars{0}\cos\pars{x} + {\rm y}'\pars{0}\sin\pars{x} + \sin\pars{x}\tan\pars{x} - \cos\pars{x}\bracks{-\,{1 \over 2\cos^{2}\pars{x}} + \half} \end{align} $$\color{#0000ff}{\large% {\rm y}\pars{x} = \bracks{{\rm y}\pars{0} - \half}\cos\pars{x} + {\rm y}'\pars{0}\sin\pars{x} + \sin\pars{x}\tan\pars{x} + \half\sec\pars{x}} $$


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