Tutorial Practice

$$y^{\prime \prime }+11y^{\prime }+24y=0$$ $$k^2+11k+24=0=(k+3)(k+8):k=-8,-3\text{ (real distinct roots)}$$ $$y=C_1{e}^{-8x}+C_2{e}^{-3x}\text{ by the principle of superposition}$$

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$$y^{\prime \prime }+2y^{\prime }+y=0$$ $$k^2+2k+1=0=(k+1{)}^2:k=-1\text{ (real repeated root)}$$ $$y=C_1{e}^{-x}+C_{2}x{e}^{-x}\text{ by the principle of superposition}$$

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$$y^{\prime \prime }+5y^{\prime }+8y=0$$ $$k^2+5k+8=0:k=-\frac{5}{2}\pm i\frac{\sqrt{7}}{2}\text{ (complex conjugate roots)}$$ $$y=e^{\frac{-5x}{2}}\left(c_1\cos \frac{\sqrt{7}x}{2}+c_2\sin \frac{\sqrt{7}x}{2}\right)$$

Note: there are actually complex parts of the solution, but they’re ignored in favour of practicality.

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$$y^{\prime \prime }+3y^{\prime }+2y=x^3+2x$$ $$y=y_{\text{complimentary}}+y_{\text{particular}}$$ $$y_{\text{complimentary}}:k^2+3k+2=(k+2)(k+1)=0:k=-2,-1$$ $$\therefore y_{\text{complimentary}}=y_c=c_1{e}^{-x}+c_2{e}^{-2x}$$ $$y_{\text{particular}}=y_p:A{x}^3+B{x}^2+Cx+D$$

Note: look into other possible forms, see where it comes from actually

$$y_p^{\prime }=3A{x}^2+2Bx+C$$ $$y_p^{\prime \prime }=6Ax+2B$$ $$\begin{array}{rl}{x}^3+2x& =(6Ax+2B)+3(3A{x}^2+2Bx+C)+2(A{x}^3+B{x}^2+Cx+D)\\ & =(2A)x^3+(9A+2B)x^2+(6A+6B+2C)x+(2B+3C+2D)\end{array}$$ $$\{\begin{array}{l}2A=1\\ 9A+2B=0\\ 6A+6B+2C=2\\ 2B+3C+2D=0\end{array}$$ $$\left[\begin{array}{cccccc}2& 0& 0& 0& |& 1\\ 9& 2& 0& 0& |& 0\\ 6& 6& 2& 0& |& 2\\ 0& 2& 3& 2& |& 0\end{array}\right]$$ $$\{\begin{array}{l}A=\frac{1}{2}\\ B=-\frac{9}{4}\\ C=\frac{25}{4}\\ D=-\frac{57}{8}\end{array}$$ $$y=c_1{e}^{-x}+c_2{e}^{-2x}+\frac{1}{2}{x}^3-\frac{9}{4}{x}^2+\frac{25}{4}x-\frac{57}{8}$$