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Question Number 130849 by TITA last updated on 29/Jan/21

Commented by Ñï= last updated on 29/Jan/21

Answered by EDWIN88 last updated on 30/Jan/21

$$-2y=-2a\cos 2x-2b\sin 2x$$$$y^{\prime }=-2a\sin 2x+2b\cos 2x$$$$\underset{―}{y^{″}=-4a\cos 2x-4b\sin 2x}+$$$$(-6a+2b)\cos 2x+(-2a-6b)\sin 2x=6\cos 2x+0\sin 2x$$$$\{\begin{array}{l}-2a-6b=0\Rightarrow a=-3b\\ -6a+2b=6;b=\frac{3}{10}\wedge a=-\frac{9}{10}\end{array}$$$$particularsolution:$$$$y_p=-\frac{9}{10}\cos 2x+\frac{3}{10}\sin 2x$$$$$$

Answered by mathmax by abdo last updated on 30/Jan/21

$${\mathrm{xy}}^2{\mathrm{y}}^{{}^{\prime }}-{2\mathrm{y}}^3=\mathrm{x}\mathrm{let}\mathrm{z}={\mathrm{y}}^3\Rightarrow {\mathrm{z}}^{{}^{\prime }}={3\mathrm{y}}^2{\mathrm{y}}^{{}^{\prime }}$$$$\mathrm{e}\Rightarrow \mathrm{x}\left(\frac{{\mathrm{z}}^{{}^{\prime }}}{3}\right)-2\mathrm{z}=\mathrm{x}\Rightarrow {\mathrm{xz}}^{{}^{\prime }}-6\mathrm{z}=3\mathrm{x}$$$$\mathrm{h}\to {\mathrm{xz}}^{{}^{\prime }}-6\mathrm{z}=0\Rightarrow {\mathrm{xz}}^{{}^{\prime }}=6\mathrm{z}\Rightarrow \frac{{\mathrm{z}}^{{}^{\prime }}}{\mathrm{z}}=\frac{6}{\mathrm{x}}\Rightarrow \mathrm{lnz}=6\ln \mid \mathrm{x}\mid +\mathrm{c}\Rightarrow \mathrm{z}=\mathrm{k}{\mathrm{x}}^6$$$$\mathrm{mvc}\mathrm{method}\to {\mathrm{z}}^{{}^{\prime }}={\mathrm{k}}^{{}^{\prime }}{\mathrm{x}}^6+6\mathrm{k}{\mathrm{x}}^5$$$$\mathrm{e}\to {\mathrm{k}}^{{}^{\prime }}{\mathrm{x}}^7+{6\mathrm{kx}}^6-{6\mathrm{kx}}^6=3\mathrm{x}\Rightarrow {\mathrm{k}}^{{}^{\prime }}=\frac{3}{{\mathrm{x}}^6}\Rightarrow \mathrm{k}=3\int \frac{\mathrm{dx}}{{\mathrm{x}}^6}=3\int {\mathrm{x}}^{-6}\mathrm{dx}$$$$=3\times \frac{1}{-5}{\mathrm{x}}^{-5}+\lambda =-\frac{3}{{5\mathrm{x}}^5}+\lambda \Rightarrow \mathrm{z}\left(\mathrm{x}\right)=(-\frac{3}{{5\mathrm{x}}^5}+\lambda ){\mathrm{x}}^6$$$$=-\frac{3}{5}\mathrm{x}+\lambda {\mathrm{x}}^6\Rightarrow \mathrm{y}{=}^3\sqrt{\mathrm{z}\left(\mathrm{x}\right)}{=}^3\sqrt{\lambda {\mathrm{x}}^6-\frac{3}{5}\mathrm{x}}$$

Answered by mathmax by abdo last updated on 30/Jan/21

$$2){\mathrm{y}}^{{}^{″}}+\mathrm{y}-2=6\cos \left(2\mathrm{x}\right)$$$$\mathrm{h}\to {\mathrm{r}}^2+\mathrm{r}-2=0\to \mathrm{\Delta }=9\Rightarrow {\mathrm{r}}_1=\frac{-1+3}{2}=1\mathrm{and}{\mathrm{r}}_2=\frac{-1-3}{2}=-2\Rightarrow$$$${\mathrm{y}}_{\mathrm{h}}={\mathrm{ae}}^{\mathrm{x}}+{\mathrm{be}}^{-2\mathrm{x}}={\mathrm{au}}_1+{\mathrm{bu}}_2$$$$\mathrm{W}({\mathrm{u}}_1,{\mathrm{u}}_2)=\left|\begin{array}{c}{\mathrm{e}}^{\mathrm{x}}{\mathrm{e}}^{-2\mathrm{x}}\\ {\mathrm{e}}^{\mathrm{x}}-{2\mathrm{e}}^{-2\mathrm{x}}\end{array}\right|=-{2\mathrm{e}}^{-\mathrm{x}}-{\mathrm{e}}^{-\mathrm{x}}=-{3\mathrm{e}}^{-\mathrm{x}}\ne 0$$$${\mathrm{W}}_1=\left|\begin{array}{c}\mathrm{o}{\mathrm{e}}^{-2\mathrm{x}}\\ 6\cos \left(2\mathrm{x}\right)-{2\mathrm{e}}^{-2\mathrm{x}}\end{array}\right|=-{6\mathrm{e}}^{-2\mathrm{x}}\cos \left(2\mathrm{x}\right)$$$${\mathrm{W}}_2=\left|\begin{array}{c}{\mathrm{e}}^{-2\mathrm{x}}0\\ -{2\mathrm{e}}^{-\mathrm{x}}6\cos \left(2\mathrm{x}\right)\end{array}\right|={6\mathrm{e}}^{-2\mathrm{x}}\cos \left(2\mathrm{x}\right)$$$${\mathrm{V}}_1=\int \frac{{\mathrm{W}}_1}{\mathrm{W}}\mathrm{dx}=-6\int \frac{{\mathrm{e}}^{-2\mathrm{x}}\cos \left(2\mathrm{x}\right)}{-{3\mathrm{e}}^{-\mathrm{x}}}\mathrm{dx}=2\int {\mathrm{e}}^{-\mathrm{x}}\cos \left(2\mathrm{x}\right)\mathrm{dx}$$$$=2\mathrm{Re}(\int {\mathrm{e}}^{-\mathrm{x}+2\mathrm{ix}}\mathrm{dx})\mathrm{and}\int {\mathrm{e}}^{(-1+2\mathrm{i})\mathrm{x}}\mathrm{dx}=\frac{1}{-1+2\mathrm{i}}{\mathrm{e}}^{(-1+2\mathrm{i})\mathrm{x}}$$$$=-\frac{1}{1-2\mathrm{i}}{\mathrm{e}}^{-\mathrm{x}}(\cos \left(2\mathrm{x}\right)+\mathrm{isin}\left(2\mathrm{x}\right))=-\frac{{\mathrm{e}}^{-\mathrm{x}}}{5}(1+2\mathrm{i})(\cos \left(2\mathrm{x}\right)+\mathrm{isin}\left(2\mathrm{x}\right))$$$$=-\frac{{\mathrm{e}}^{-\mathrm{x}}}{5}\{\cos \left(2\mathrm{x}\right)+\mathrm{isin}\left(2\mathrm{x}\right)+2\mathrm{icos}\left(2\mathrm{x}\right)-2\sin \left(2\mathrm{x}\right)\}\Rightarrow$$$${\mathrm{v}}_1=-\frac{2}{5}{\mathrm{e}}^{-\mathrm{x}}(\cos \left(2\mathrm{x}\right)-2\sin \left(2\mathrm{x}\right))$$$${\mathrm{V}}_2=\int \frac{{\mathrm{W}}_2}{\mathrm{W}}\mathrm{dx}=\int \frac{{6\mathrm{e}}^{-2\mathrm{x}}\cos \left(2\mathrm{x}\right)\mathrm{dx}}{-{3\mathrm{e}}^{-\mathrm{x}}}\mathrm{dx}$$$$=-2\int {\mathrm{e}}^{-\mathrm{x}}\cos \left(2\mathrm{x}\right)\mathrm{dx}=\frac{2}{5}{\mathrm{e}}^{-\mathrm{x}}(\cos \left(2\mathrm{x}\right)-2\sin \left(2\mathrm{x}\right))\Rightarrow$$$${\mathrm{y}}_{\mathrm{p}}={\mathrm{u}}_1{\mathrm{v}}_1+{\mathrm{u}}_2{\mathrm{v}}_2=-\frac{2}{5}(\cos \left(2\mathrm{x}\right)-2\sin \left(2\mathrm{x}\right))+\frac{2}{5}{\mathrm{e}}^{-3\mathrm{x}}(\cos \left(2\mathrm{x}\right)-2\sin \left(2\mathrm{x}\right))$$$$=(-\frac{2}{5}+\frac{2}{5}{\mathrm{e}}^{-3\mathrm{x}})\cos \left(2\mathrm{x}\right)+(\frac{4}{5}-\frac{4}{5}{\mathrm{e}}^{-3\mathrm{x}})\sin \left(2\mathrm{x}\right)$$$$\mathrm{general}\mathrm{solution}\mathrm{is}$$$$\mathrm{y}={\mathrm{y}}_{\mathrm{h}}+{\mathrm{y}}_{\mathrm{p}}={\mathrm{ae}}^{\mathrm{x}}+{\mathrm{be}}^{-2\mathrm{x}}+(-\frac{2}{5}+\frac{2}{5}{\mathrm{e}}^{-3\mathrm{x}})\cos \left(2\mathrm{x}\right)$$$$+\frac{4}{5}(1-{\mathrm{e}}^{-3\mathrm{x}})\sin \left(2\mathrm{x}\right)$$

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