# Determine the complete real solution to a homogeneous differential equation

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

y^{4}-16y=u^{1}+u
y′′′′−16y=u′+u


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.

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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

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.

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