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} $$\color{#0000ff}{\large% {\rm y}\pars{x} = \bracks{{\rm y}\pars{0} - \half}\cos\pars{x} + {\rm y}'\pars{0}\sin\pars{x} + \sin\pars{x}\tan\pars{x} + \half\sec\pars{x}} $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.