Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Hello i am having some problems working out how to attack this assignment, and after spending hours on it, have i resolved to ask you guys here for help.

I have been given the following differential equation


And i have been asked to determine the complete real solution. I just completely stuck on this assignment and hope that one of you guys can give me a push in the right direction.

share|cite|improve this question
Solve y in terms of u or u in terms of y? – kennytm Sep 5 '10 at 16:55
Just to check: $y^{(4)}-16y=u^{\prime}+u$ ? – J. M. Sep 5 '10 at 16:57
Yes to J. M. Hmm i not completly sure what it is called in english, but in danish is it "Bestem den fuldstændige reelle løsning to den homogene differentialligning" – DoomStone Sep 5 '10 at 17:04
And the answer to Kenny's question, should it be y as a function of u or vice-versa? – J. M. Sep 5 '10 at 17:14
y as a function – DoomStone Sep 5 '10 at 17:15
up vote 0 down vote accepted

Well, you could use Variation of Parameters.

The fourth order ODE $$ y^{(4)} - 16 y = u' + u $$ has the homogeneous solution $$ y_h(x) = c_1 \cosh(2 x) + c_2 \sinh(2 x) + c_3 \cos(2 x) +c_4 \sin(2 x). $$ Now, for the particular solution, take $$ y_p(x) = a_1(x) \cosh(2 x) + a_2(x) \sinh(2 x) + a_3(x) \cos(2 x) + a_4(x) \sin(2 x). $$ where \begin{align} a_1'(x) \cosh(2 x) + a_2'(x) \sinh(2 x) + a_3'(x) \cos(2 x) + a_4'(x) \sin(2 x) &=0 \\ a_1'(x) \sinh(2 x) + a_2'(x) \cosh(2 x) - a_3'(x) \sin(2 x) + a_4'(x) \cos(2 x)&=0 \\ a_1'(x) \cosh(2 x) + a_2'(x) \sinh(2 x) - a_3'(x) \cos(2 x) - a_4'(x) \sin(2 x)&=0 \end{align}

Substituting in the ODE we have $$ a_1'(x) \sinh(2 x) + a_2'(x) \cosh(2 x) + a_3'(x) \sin(2 x) - a_4'(x) \cos(2 x) = \frac{1}{8}\big(u'(x) +u(x)\big). $$ This four equations can be solved for $a_1'(x)$, $a_2'(x)$,$a_3'(x)$ and $a_4'(x)$, leading to \begin{align} a_1'(x) &= -\frac{\sinh(2 x)}{16} \big(u'(x) + u(x)\big)\\ a_2'(x) &= \frac{\cosh(2 x)}{16} \big(u'(x) + u(x)\big)\\ a_3'(x) &= \frac{\sin(2 x)}{16} \big(u'(x) + u(x)\big)\\ a_4'(x) &= -\frac{\cos(2 x)}{16} \big(u'(x) + u(x)\big) \end{align}

and, assuming that it's an Initial Value Problem given at $x = 0$ (w.l.g), \begin{align} a_1(x) &= -\frac{\sinh(2 x) u(x)}{16} - \int_0^x \frac{\sinh(2 t) - 2\cosh(2 t)}{16} u(t)\, dt\\ a_2(x) &= \frac{\cosh(2 x) u(x) - u(0)}{16} + \int_0^x \frac{\cosh(2 t) - 2\sinh(2 t)}{16} u(t)\, dt\\ a_3(x) &= \frac{\sin(2 x) u(x)}{16} + \int_0^x \frac{\sin(2 t) - 2 \cos(2 t)}{16} u(t)\, dt\\ a_4(x) &= -\frac{\cos(2 x) u(x) - u(0)}{16} - \int_0^x \frac{\cos(2 t) + 2\sin(2 t)}{16} u(t)\, dt \end{align}

the solution is \begin{multline} y(x) = \big(c_1 + a_1(x)\big) \cosh (2 x) + \big(c_2 + a_2(x)\big) \sinh (2 x) \\ + \big(c_3 + a_3(x)\big) \cos (2 x) + \big(c_4 + a_4(x)\big) \sin (2 x). \end{multline}

For a Boundary Value Problem the construction is a bit different, but the same principle can be applied.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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