# Solving a Fourth-Order Linear Homogeneous Differential Equation

I like to solve the ordinary fourth order homogeneous differential equation given by

$$\displaystyle \frac{d^{4}\theta}{d z^{4}} + \lambda \cdot \theta = 0$$

with a constant coefficient $$\lambda$$. Using the characteristic equation, I come up with the following roots

$$r_1=-\sqrt[4]{-\lambda}, r_2=\sqrt[4]{-\lambda}, r_3=-i\sqrt[4]{-\lambda}, r_4=i\sqrt[4]{-\lambda}$$

By putting the roots into the general solutions you get the following solution

$$\displaystyle \theta{\left(z \right)} = C_{1} e^{- z \sqrt[4]{- \lambda}} + C_{2} e^{z \sqrt[4]{- \lambda}} + C_{3} sin\left(z \sqrt[4]{- \lambda}\right) + C_{4} cos\left(z \sqrt[4]{- \lambda}\right)$$

(Here is already the first problem, I am not sure if the solution is correct).

For the specific solution only two boundary conditions are given: $$\theta(0)=C, \theta(z \rightarrow \infty)=0$$

Assuming $$C<0$$, the authors come up with the following solution

$$\theta(z) = C e^{\left(-z \sqrt[4]{4\lambda)} \right)} cos(z \sqrt[4]{4 \lambda})$$.

I don't understand how to obtain this solution based on the boundary conditions. Nor can I understand where the 4 under the root comes from. I am grateful for any help. Thank you!

• Are we assuming $\lambda > 0$? Apr 15 at 13:25

I think you have a slight error here. It should be $$\lambda/4$$, not $$4\lambda$$. It comes from the fact that $$\sqrt[4]{-1} = \frac{\pm 1 \pm i}{\sqrt{2}}.$$ So the values for $$r$$ are $$r = \sqrt[4]{-\lambda} = \frac{\pm1\pm i}{\sqrt{2}}\sqrt[4]{\lambda} = (\pm 1\pm i)\sqrt[4]{\frac{\lambda}{4}},$$ which means the solutions are $$Ce^{\pm z\sqrt[4]{\lambda/4}}\cos\left(z\sqrt[4]{\frac{\lambda}{4}}\right), Ce^{\pm z\sqrt[4]{\lambda/4}}\sin\left(z\sqrt[4]{\frac{\lambda}{4}}\right).$$ The constraint $$\theta(z\rightarrow \infty) = 0$$ forces the exponential to have a negative argument, and $$\theta(0) = C$$ sets the coefficient of the cosine term. However, this doesn't say anything about the coefficient of the sine term, so the most general solution is $$\theta(z) = Ce^{- z\sqrt[4]{\lambda/4}}\cos\left(z\sqrt[4]{\frac{\lambda}{4}}\right) + Ke^{- z\sqrt[4]{\lambda/4}}\sin\left(z\sqrt[4]{\frac{\lambda}{4}}\right)$$ for some constant $$K$$.
• That was really helpful, thank you! Only one thing is still unclear to me. The problem has a characteristic equation with complex conjugate roots ($a+bi$), resulting in the general solution \begin{align} e^{az}(C_1 cos(bz) + C_2 sin(bz)) \end{align} rather than \begin{align} Ce^{\pm (a+bi)z}\cos ((a+bi)z) \end{align} How do you come up with the second solution? Apr 15 at 17:09
• @tsauter You make linear combinations of the complex exponentials. We have $\exp[(a + bi)z] + \exp[(a - bi)z] = 2\exp(az)\cos(bz)$ and $\exp[(a + bi)z] - \exp[(a - bi)z] = 2i\exp(az)\sin(bz)$. Apr 16 at 3:49