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Question Number 126408 by morarupaula last updated on 20/Dec/20

Answered by Dwaipayan Shikari last updated on 20/Dec/20

$$\stackrel{\bullet \bullet \bullet }{x}+\stackrel{\bullet }{x}=0x=e^{\lambda t}$$$${\lambda }^3+\lambda =0\Rightarrow \lambda =0,\lambda =\pm i$$$$x=\mathrm{\Lambda }+\mathrm{\Gamma }{e}^{\lambda ti}+\mathrm{\Phi }{e}^{-\lambda it}=\mathrm{\Lambda }+\mathrm{\Psi }cos\left(t\right)+\beta sin\left(t\right)$$

Answered by Olaf last updated on 22/Dec/20

$$x^{{}^{‴}}+x^{\prime }=0$$$$x^{‴}{x}^{″}+x^{″}{x}^{\prime }=0$$$$\frac{1}{2}{x}^{{}^{″2}}+\frac{1}{2}{x}^{\prime 2}={\mathrm{C}}_1({\mathrm{C}}_1⩾0)$$$$x^{{}^{″2}}+x^{\prime 2}={\mathrm{C}}_2({\mathrm{C}}_2⩾0)$$$$\frac{x^{″2}}{\sqrt{{\mathrm{C}}_2-x^{\prime 2}}}=1$$$$\arcsin \left(\frac{x^{\prime }}{\sqrt{{\mathrm{C}}_2}}\right)=t+{\mathrm{C}}_3$$$$\arcsin \left(\frac{x^{\prime }}{{\mathrm{C}}_4}\right)=t+{\mathrm{C}}_3$$$$x^{\prime }={\mathrm{C}}_4\sin (t+{\mathrm{C}}_3)$$$$x=-{\mathrm{C}}_4\cos (t+{\mathrm{C}}_3)+{\mathrm{C}}_5$$$$\mathrm{General}\mathrm{solution}:$$$$x=\alpha \cos (t+\beta )+\gamma$$

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