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Question Number 103478 by mohammad17 last updated on 15/Jul/20

Answered by bemath last updated on 15/Jul/20

$$HE\Rightarrow {\lambda }^2+3\lambda -2=0$$$${(\lambda +\frac{3}{2})}^2-\frac{9}{4}-2=0$$$${(\lambda +\frac{3}{2})}^2=\frac{17}{4}\Rightarrow \lambda =-\frac{3}{2}\pm \frac{\sqrt{17}}{2}$$$$y_h=C_1{e}^{\frac{(-3+\sqrt{17})x}{2}}+C_2{e}^{\frac{(-3-\sqrt{17})x}{2}}$$$$particularsolution$$$$y_p=-\frac{11}{101}\sin 3t-\frac{9}{101}\cos 3t$$$$generalsolution$$$$y=C_1{e}^{\frac{(-3+\sqrt{17})x}{2}}+C_2{e}^{\frac{(-3-\sqrt{17})x}{2}}-$$$$\frac{11}{101}\sin 3t-\frac{9}{101}\cos 3t$$$$y\left(0\right)\Rightarrow C_1+C_2=\frac{9}{101}$$$$y^{\prime }\left(x\right)=\left(\frac{-3+\sqrt{17}}{2}\right)C_1{e}^{\frac{(-3+\sqrt{17})x}{2}}-$$$$C_2\left(\frac{3+\sqrt{17}}{2}\right)e^{\frac{(-3-\sqrt{17})x}{2}}$$

Answered by Worm_Tail last updated on 15/Jul/20

$$\frac{d^{2}y}{{dt}^2}+3\frac{dy}{dt}-2y=2sin3t$$$$L(\frac{d^{2}y}{{dt}^2}+3\frac{dy}{dt}-2y)=L\left(2sin3t\right)$$$$L\left(\frac{d^{2}y}{{dt}^2}\right)+L\left(3\frac{dy}{dt}\right)-L\left(2y\right)=\frac{6}{s^2+9}$$$$s^{2}F\left(s\right)-sy\left(0\right)-y^{\prime }\left(0\right)+3(sF\left(s\right)-y\left(0\right))-2F\left(s\right)=\frac{6}{s^2+9}$$$$s^{2}F\left(s\right)-(-2)+3sF\left(s\right)-2F\left(s\right)=\frac{6}{s^2+9}$$$$F\left(s\right)[s^2+3s-2]=\frac{6}{s^2+9}-2$$$$F\left(s\right)[s^2+3s-2]=\frac{-2s^2-12}{s^2+9}$$$$F\left(s\right)=\frac{-2s^2-12}{(s^2+9)(s^2+3s-2)}$$$$F\left(s\right)=\frac{9}{101}\left(\frac{s}{{(s-\frac{3}{2})}^2-\frac{19}{4}}\right)-\frac{142}{101}\left(\frac{1}{{(s-\frac{3}{2})}^2-\frac{19}{4}}\right)-\frac{9}{101}\left(\frac{s}{s^2+9}\right)-\frac{33}{101}\left(\frac{1}{s^2+9}\right)$$$$evaluating{L}^{-1}ofbothsides$$$$f\left(t\right)=\frac{e^{\frac{-t(3+\sqrt{17)}}{2}}}{34\left(101\right)}(-311\sqrt{17}{e}^{t\sqrt{17}}+153e^{t\sqrt{17}}+153+311\sqrt{17})-\frac{(9x+11)sin\left(3t\right)}{3\left(101\right)}$$

Answered by mathmax by abdo last updated on 15/Jul/20

$${\mathrm{y}}^{{}^{″}}+{3\mathrm{y}}^{{}^{\prime }}-2\mathrm{y}=2\sin \left(3\mathrm{t}\right)$$$$\mathrm{he}\to {\mathrm{y}}^{{}^{″}}+{3\mathrm{y}}^{{}^{\prime }}-2\mathrm{y}=0\to {\mathrm{r}}^2+3\mathrm{r}-2=0\to \mathrm{\Delta }=9+8=17\Rightarrow$$$${\mathrm{r}}_1=\frac{-3+\sqrt{17}}{2}\mathrm{and}{\mathrm{r}}_2=\frac{-3-\sqrt{17}}{2}\Rightarrow {\mathrm{y}}_{\mathrm{h}}={\mathrm{ae}}^{{\mathrm{r}}_1\mathrm{t}}+\mathrm{b}{\mathrm{e}}^{{\mathrm{r}}_2\mathrm{t}}={\mathrm{au}}_1+{\mathrm{bu}}_2$$$$\mathrm{W}({\mathrm{u}}_1,{\mathrm{u}}_2)=\left|\begin{array}{c}{\mathrm{e}}^{{\mathrm{r}}_1\mathrm{t}}{\mathrm{e}}^{{\mathrm{r}}_2\mathrm{t}}\\ {\mathrm{r}}_1{\mathrm{e}}^{{\mathrm{r}}_1\mathrm{t}}{\mathrm{r}}_2{\mathrm{e}}^{{\mathrm{r}}_2\mathrm{t}}\end{array}\right|=({\mathrm{r}}_2-{\mathrm{r}}_1){\mathrm{e}}^{({\mathrm{r}}_1+{\mathrm{r}}_2)\mathrm{t}}=-\sqrt{17}{\mathrm{e}}^{-3\mathrm{t}}\ne 0$$$${\mathrm{W}}_1=\left|\begin{array}{c}\mathrm{o}{\mathrm{e}}^{{\mathrm{r}}_2\mathrm{t}}\\ 2\sin \left(3\mathrm{t}\right){\mathrm{r}}_2{\mathrm{e}}^{{\mathrm{r}}_2\mathrm{t}}\end{array}\right|=-{2\mathrm{e}}^{{\mathrm{r}}_2\mathrm{t}}\sin \left(3\mathrm{t}\right)$$$${\mathrm{W}}_2=\left|\begin{array}{c}{\mathrm{e}}^{{\mathrm{r}}_1\mathrm{t}}0\\ {\mathrm{r}}_1{\mathrm{e}}^{{\mathrm{r}}_1\mathrm{t}}2\sin \left(3\mathrm{t}\right)\end{array}\right|={2\mathrm{e}}^{\mathrm{r}1\mathrm{t}}\sin \left(3\mathrm{t}\right)$$$${\mathrm{v}}_1=\int \frac{{\mathrm{w}}_1}{\mathrm{w}}\mathrm{dt}=-\int \frac{{2\mathrm{e}}^{{\mathrm{r}}_2\mathrm{t}}\sin \left(3\mathrm{t}\right)}{-\sqrt{17}{\mathrm{e}}^{-3\mathrm{t}}}=\frac{2}{\sqrt{17}}\int {\mathrm{e}}^{(3+{\mathrm{r}}_2)\mathrm{t}}\sin \left(3\mathrm{t}\right)\mathrm{dt}$$$$=\frac{2}{\sqrt{17}}\int {\mathrm{e}}^{\left(\frac{3-\sqrt{17}}{2}\right)\mathrm{t}}\sin \left(3\mathrm{t}\right)\mathrm{dt}=\frac{2}{\sqrt{17}}\mathrm{Im}(\int {\mathrm{e}}^{(\frac{3-\sqrt{17}}{2}+3\mathrm{i})\mathrm{t}}\mathrm{dt})$$$$\mathrm{and}\int {\mathrm{e}}^{(...)\mathrm{t}}\mathrm{dt}=\frac{1}{(\frac{3-\sqrt{17}}{2}+3\mathrm{i})}{\mathrm{e}}^{(\frac{3-\sqrt{17}}{2}+3\mathrm{i})\mathrm{t}}$$$$=\frac{2}{(3-\sqrt{17}+6\mathrm{i})}{\mathrm{e}}^{(...)}=\frac{2(3-\sqrt{17}-6\mathrm{i})}{{(3-\sqrt{17})}^2+36}{\mathrm{e}}^{\frac{3-\sqrt{17}}{2}\mathrm{t}}\{\cos \left(3\mathrm{t}\right)+\mathrm{isin}\left(3\mathrm{t}\right)\}\mathrm{rest}$$$$\mathrm{to}\mathrm{extract}\mathrm{Im}(\int ..)..$$$${\mathrm{v}}_2=\int \frac{{\mathrm{w}}_2}{\mathrm{w}}\mathrm{dt}=\int \frac{{2\mathrm{e}}^{{\mathrm{r}}_1\mathrm{t}}\sin \left(3\mathrm{t}\right)}{-\sqrt{17}{\mathrm{e}}^{-3\mathrm{t}}}=-\frac{2}{\sqrt{17}}\int {\mathrm{e}}^{(3+{\mathrm{r}}_1)\mathrm{t}}\sin \left(3\mathrm{t}\right)$$$$=-\frac{2}{\sqrt{17}}\mathrm{Im}(\int {\mathrm{e}}^{(3+{\mathrm{r}}_1+3\mathrm{i})\mathrm{t}}\mathrm{dt})=....\mathrm{we}\mathrm{follow}\mathrm{the}\mathrm{same}\mathrm{way}..$$$$\Rightarrow {\mathrm{y}}_{\mathrm{p}}={\mathrm{u}}_1{\mathrm{v}}_1+{\mathrm{u}}_2{\mathrm{v}}_2\mathrm{and}\mathrm{y}={\mathrm{y}}_{\mathrm{h}}+{\mathrm{y}}_{\mathrm{p}}$$

Answered by mathmax by abdo last updated on 15/Jul/20

$$\mathrm{let}\mathrm{use}\mathrm{laplace}\mathrm{transform}$$$${\mathrm{y}}^{{}^{″}}+{3\mathrm{y}}^{{}^{\prime }}-2\mathrm{y}=2\sin \left(3\mathrm{t}\right)\Rightarrow \mathrm{L}\left({\mathrm{y}}^{{}^{″}}\right)+3\mathrm{L}\left({\mathrm{y}}^{{}^{\prime }}\right)-2\mathrm{L}\left(\mathrm{y}\right)=2\mathrm{L}\left(\sin \left(3\mathrm{t}\right)\right)\Rightarrow$$$${\mathrm{t}}^2\mathrm{L}\left(\mathrm{y}\right)-\mathrm{ty}\left(\mathrm{o}\right)-{\mathrm{y}}^{{}^{\prime }}\left(\mathrm{o}\right)+3(\mathrm{t}\mathrm{L}\left(\mathrm{y}\right)-\mathrm{y}\left(\mathrm{o}\right))-2\mathrm{L}\left(\mathrm{y}\right)=2\mathrm{L}\left(\sin \left(3\mathrm{t}\right)\right)\Rightarrow$$$$({\mathrm{t}}^2+3\mathrm{t}-2)\mathrm{L}\left(\mathrm{y}\right)+2=2\mathrm{L}\left(\sin \left(3\mathrm{t}\right)\right)\mathrm{and}$$$$\mathrm{L}\left(\sin \left(3\mathrm{t}\right)\right)={\int }_0^{\mathrm{\infty }}\sin \left(3\mathrm{x}\right){\mathrm{e}}^{-\mathrm{tx}}\mathrm{dx}=\mathrm{Im}\left({\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{tx}+3\mathrm{ix}}\mathrm{dx}\right)$$$$=\mathrm{Im}\left({\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{(-\mathrm{t}+3\mathrm{i})\mathrm{x}}\mathrm{dx}\right)\mathrm{and}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{(-\mathrm{t}+3\mathrm{i})\mathrm{x}}\mathrm{dx}={\left[\frac{1}{-\mathrm{t}+3\mathrm{i}}{\mathrm{e}}^{(-\mathrm{t}+3\mathrm{i})\mathrm{x}}\right]}_0^{\mathrm{\infty }}$$$$=-\frac{1}{-\mathrm{t}+3\mathrm{i}}=\frac{1}{\mathrm{t}-3\mathrm{i}}=\frac{\mathrm{t}+3\mathrm{i}}{{\mathrm{t}}^2+9}\Rightarrow \mathrm{L}\left(\sin \left(3\mathrm{t}\right)\right)=\frac{3}{{\mathrm{t}}^2+9}$$$$\mathrm{e}\Rightarrow ({\mathrm{t}}^2+3\mathrm{t}-2)\mathrm{L}\left(\mathrm{y}\right)=-2+\frac{6}{{\mathrm{t}}^2+9}\Rightarrow \mathrm{L}\left(\mathrm{y}\right)=\frac{-2}{{\mathrm{t}}^2+3\mathrm{t}-2}+\frac{6}{({\mathrm{t}}^2+9)({\mathrm{t}}^2+3\mathrm{t}-2)}$$$$\Rightarrow \mathrm{y}\left(\mathrm{t}\right)=-{2\mathrm{L}}^{-1}\left(\frac{1}{{\mathrm{t}}^2+3\mathrm{t}-2}\right)+{6\mathrm{L}}^{-1}\left(\frac{1}{({\mathrm{t}}^2+9)({\mathrm{t}}^2+3\mathrm{t}-2)}\right)$$$$\mathrm{let}\mathrm{decompose}\mathrm{f}\left(\mathrm{t}\right)=\frac{1}{{\mathrm{t}}^2+3\mathrm{t}-2}$$$$\mathrm{\Delta }=9+8=17\Rightarrow {\mathrm{t}}_1=\frac{-3+\sqrt{17}}{2}\mathrm{and}{\mathrm{t}}_2=\frac{-3-\sqrt{17}}{2}\Rightarrow$$$$\mathrm{f}\left(\mathrm{t}\right)=\frac{1}{(\mathrm{t}-{\mathrm{t}}_1)(\mathrm{t}-{\mathrm{t}}_2)}=\frac{1}{\sqrt{17}}(\frac{1}{\mathrm{t}-{\mathrm{t}}_1}-\frac{1}{\mathrm{t}-{\mathrm{t}}_2})\Rightarrow$$$${\mathrm{L}}^{-1}\left(\mathrm{f}\right)=\frac{1}{\sqrt{17}}{\mathrm{e}}^{{\mathrm{t}}_1\mathrm{t}}-\frac{1}{\sqrt{17}}{\mathrm{e}}^{{\mathrm{t}}_2\mathrm{t}}\mathrm{let}\mathrm{decompose}\mathrm{g}\left(\mathrm{t}\right)=\frac{1}{({\mathrm{t}}^2+9)({\mathrm{t}}^2+3\mathrm{t}-2)}$$$$\Rightarrow \mathrm{g}\left(\mathrm{t}\right)=\frac{1}{(\mathrm{t}-3\mathrm{i})(\mathrm{t}+3\mathrm{i})(\mathrm{t}-{\mathrm{t}}_1)(\mathrm{t}-{\mathrm{t}}_2)}=\frac{\mathrm{a}}{\mathrm{t}-3\mathrm{i}}+\frac{\mathrm{b}}{\mathrm{t}+3\mathrm{i}}+\frac{\mathrm{c}}{\mathrm{t}-{\mathrm{t}}_1}+\frac{\mathrm{d}}{\mathrm{t}-{\mathrm{t}}_2}$$$$\mathrm{eazy}\mathrm{to}\mathrm{find}\mathrm{this}\mathrm{coefficients}\Rightarrow$$$${\mathrm{L}}^{-1}\left(\mathrm{g}\right)=\mathrm{a}{\mathrm{e}}^{3\mathrm{it}}+\mathrm{b}{\mathrm{e}}^{-3\mathrm{it}}+\mathrm{c}{\mathrm{e}}^{{\mathrm{t}}_1\mathrm{t}}+{\mathrm{de}}^{{\mathrm{t}}_2\mathrm{t}}$$$$\Rightarrow \mathrm{at}\mathrm{form}\alpha \cos \left(3\mathrm{t}\right)+\beta \sin \left(3\mathrm{t}\right)+{\mathrm{ce}}^{{\mathrm{t}}_1\mathrm{t}}+\mathrm{d}{\mathrm{e}}^{{\mathrm{t}}_2\mathrm{t}}\Rightarrow$$$$\mathrm{y}\left(\mathrm{t}\right)=\alpha \cos \left(3\mathrm{t}\right)+\beta \sin \left(3\mathrm{t}\right)+\mathrm{a}{\mathrm{e}}^{\left(\frac{-3+\sqrt{17}}{2}\right)\mathrm{t}}+\mathrm{b}{\mathrm{e}}^{\left(\frac{-3-\sqrt{17}}{2}\right)\mathrm{t}}$$$$$$

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