Homogeneous linear equations of order n with constant coefficients

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. $$\lambda=\dfrac{3\pm\sqrt{9-4\cdot2\cdot4}}{4}=\dfrac{3\pm\sqrt{-23}}{4}=\dfrac{3}{4}\pm\dfrac{\sqrt{23}}{4}i$$ Therefore we have one complex root and its conjugate. Therefore, the solution functions are: $$y_1(x)=e^{\frac{3}{4}}\cos(\dfrac{\sqrt{23}}{4}x)$$ $$y_2(x)=e^{\frac{3}{4}}\sin(\dfrac{\sqrt{23}}{4}x)$$ Any solution is written as: $$y(x)=C_1\cdot e^{\frac{3}{4}}\cos(\dfrac{\sqrt{23}}{4}x)+C_2\cdot e^{\frac{3}{4}}\sin(\dfrac{\sqrt{23}}{4}x)$$

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 $y_1(x)=e^{0x}=1$, $y_2(x)=x\cdot e^{0x}=x$.

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

• $\lambda=2\pm2i$ 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 $5$: $$y(x)=C_1+C_2\cdot x+C_3\cdot e^{-\frac{1}{2}x}+C_4\cdot e^{2x}\cos(2x)+C_5\cdot e^{2x}\sin(2x)$$