# the 4th order linear ode with 3 initial conditions

I am reading a paper in which it deals with an ODE defined by $$f''''(x)+f(x)-1=0$$ for $x\in[0,+\infty)$ where $f(0)=f'(0)=0$ and $\lim_{x\to \infty}f(x)=1$. Then it says trivially we can have a 'unique' solution $$f(x)=1+\sqrt{2}e^{-t/\sqrt{2}}\cos\left(\frac t{\sqrt{2}}+\frac \pi4\right)$$

I am a bit confused. Should we have four initial conditions to have an exact solution for a forth order ODE? How this paper obtain such solution?

You may find the paper here, page 5, equation (3.3).

The other two components of the homogeneous solution are $$A·e^{t/\sqrt{2}}\cos\left(\frac t{\sqrt{2}}+\frac \pi4\right)+B·e^{t/\sqrt{2}}\sin\left(\frac t{\sqrt{2}}+\frac \pi4\right)$$ If either of the coefficients is non-zero, you get an exponentially growing function.
• ah yea... stupid me! I got it. Since they are dealing with an minimizing problem so they are forcing to take $A=B=0$. Thank you. – spatially Dec 4 '16 at 20:21
• and what if I deal with $-f^{(6)}+f-1=0$ with in addition that $f''(0)=0$? – spatially Dec 4 '16 at 20:23
• Set $g(x)=f(x)-1$, then $g''''+g=0$ is homogeneous with characteristic polynomial $λ^4+1=0$ which has roots $λ=\frac1{\sqrt2}(\pm1\pmi)$. -- The second example would have $-λ^6+1=0$, $λ=e^{ik\frac\pi3}$, $k=0,1,...,5$. -- In the homogeneous solution components $C·e^{λt}$ remember that $e^{(a+ib)t}=e^{at}·(\cos(bt)+i\sin(bt))$ and that the complex solutions of real equations come in conjugate pairs. – LutzL Dec 4 '16 at 20:28