Solve the following ODEs:
a) $$2y''-3y'+4y=0$$
b) $$2y^{(5)}-7y^{(4)}+12y'''+8y''=0$$
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. $$$\lambda=\dfrac{3\pm\sqrt{9-4\cdot2\cdot4}}{4}=\dfrac{3\pm\sqrt{-23}}{4}=\dfrac{3}{4}\pm\dfrac{\sqrt{23}}{4}i$$$ 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 $$5$$: $$$y(x)=C_1+C_2\cdot x+C_3\cdot e^{-\frac{1}{2}x}+C_4\cdot e^{2x}\cos(2x)+C_5\cdot e^{2x}\sin(2x)$$$
Solution:
a) $$y(x)=C_1\cdot e^{\frac{3}{4}}\cos(\dfrac{\sqrt{23}}{4}x)+C_2\cdot e^{\frac{3}{4}}\sin(\dfrac{\sqrt{23}}{4}x)$$
b) $$y(x)=C_1+C_2\cdot x+C_3\cdot e^{-\frac{1}{2}x}+C_4\cdot e^{2x}\cos(2x)+C_5\cdot e^{2x}\sin(2x)$$