Problems from Homogeneous linear equations of order n with constant coefficients

Development:

a) This a linear EDO of order $n$, and homogeneous with constant coefficients. Therefore, let's write its characteristical polynomial $$p(\lambda )=2{\lambda }^2-3\lambda +4$$ and let's compute its roots. $$\lambda ={\displaystyle \frac{3\pm \sqrt{9-4\cdot 2\cdot 4}}{4}}={\displaystyle \frac{3\pm \sqrt{-23}}{4}}={\displaystyle \frac{3}{4}}\pm {\displaystyle \frac{\sqrt{23}}{4}}i$$ Therefore we have one complex root and its conjugate. Therefore, the solution functions are: $$y_1(x)=e^{\frac{3}{4}}\cos ({\displaystyle \frac{\sqrt{23}}{4}}x)$$ $$y_2(x)=e^{\frac{3}{4}}\sin ({\displaystyle \frac{\sqrt{23}}{4}}x)$$ Any solution is written as: $$y(x)=C_1\cdot e^{\frac{3}{4}}\cos ({\displaystyle \frac{\sqrt{23}}{4}}x)+C_2\cdot e^{\frac{3}{4}}\sin ({\displaystyle \frac{\sqrt{23}}{4}}x)$$

b) This is the same case as the previous one. Therefore, let's write the characteristical polynomial: $$p(\lambda )=2{\lambda }^5-7{\lambda }^4+12{\lambda }^3+8{\lambda }^2={\lambda }^2(2{\lambda }^3-7{\lambda }^2+12\lambda +8)$$ Its roots are:

• $\lambda =0$ with multiplicity $2$. Therefore it gives the functions $y_1(x)=e^{0x}=1$, $y_2(x)=x\cdot e^{0x}=x$.

• $\lambda =-{\displaystyle \frac{1}{2}}$ simple. Therefore it gives the function: $y_3(x)=e^{-\frac{1}{2}x}$.

• $\lambda =2\pm 2i$ one simple root with its conjugate. Therefore they give the functions: $y_4(x)=e^{2x}\cos (2x)$,$y_5(x)=e^{2x}\sin (2x)$.

Thus, any solution is a linear combination of these $5$: $$y(x)=C_1+C_2\cdot x+C_3\cdot e^{-\frac{1}{2}x}+C_4\cdot e^{2x}\cos (2x)+C_5\cdot e^{2x}\sin (2x)$$

Solution:

a) $y(x)=C_1\cdot e^{\frac{3}{4}}\cos ({\displaystyle \frac{\sqrt{23}}{4}}x)+C_2\cdot e^{\frac{3}{4}}\sin ({\displaystyle \frac{\sqrt{23}}{4}}x)$

b) $y(x)=C_1+C_2\cdot x+C_3\cdot e^{-\frac{1}{2}x}+C_4\cdot e^{2x}\cos (2x)+C_5\cdot e^{2x}\sin (2x)$

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