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Question Number 103203 by bobhans last updated on 13/Jul/20

$$y^{″}+4y^{\prime }+4y=\frac{e^{-2x}}{x^2}$$

Answered by bramlex last updated on 14/Jul/20

$$\mathrm{characteristic}\mathrm{eq}$$$${\chi }^2+4\chi +4=0;{(\chi +2)}^2=0$$$${\mathrm{y}}_{\mathrm{h}}={\mathrm{C}}_1{\mathrm{e}}^{-2\mathrm{x}}+{\mathrm{C}}_2{\mathrm{xe}}^{-2\mathrm{x}}$$$$\mathrm{now}\mathrm{we}\mathrm{apply}\mathrm{reduction}\mathrm{of}\mathrm{order}$$$$\mathrm{set}\mathrm{y}={\mathrm{e}}^{-2\mathrm{x}}.\mathrm{z},\mathrm{substituting}\mathrm{this}$$$$\mathrm{into}\mathrm{the}\mathrm{original}\mathrm{differential}$$$$\mathrm{equation}\mathrm{yields}$$$${\mathrm{e}}^{-2\mathrm{x}}({\mathrm{z}}^{″}-{4\mathrm{z}}^{\prime }+4\mathrm{z})+{\mathrm{e}}^{-2\mathrm{x}}({\mathrm{z}}^{\prime }-2\mathrm{z})$$$$+4\mathrm{z}{\mathrm{e}}^{-2\mathrm{x}}=\frac{{\mathrm{e}}^{-2\mathrm{x}}}{{\mathrm{x}}^2}.\mathrm{which}$$$$\mathrm{simplifies}\mathrm{to}{\mathrm{z}}^{″}=\frac{1}{{\mathrm{x}}^2}$$$$\mathrm{intrgrating}\mathrm{this}\mathrm{twice}\mathrm{succession}$$$$\mathrm{yields}\mathrm{z}=-\ln \mid \mathrm{x}\mid +{\mathrm{C}}_1\mathrm{x}+{\mathrm{C}}_2$$$$\mathrm{hence}\mathrm{a}\mathrm{general}\mathrm{solution}\mathrm{is}\mathrm{given}$$$$\mathrm{by}\mathrm{y}={\mathrm{e}}^{-2\mathrm{x}}\{-\ln \mid \mathrm{x}\mid +{\mathrm{C}}_2\mathrm{x}+{\mathrm{C}}_1\}$$$$\mathrm{y}={\mathrm{C}}_1{\mathrm{e}}^{-2\mathrm{x}}+{\mathrm{C}}_2\mathrm{x}{\mathrm{e}}^{-2\mathrm{x}}-{\mathrm{e}}^{-2\mathrm{x}}\ln \mid \mathrm{x}\mid ◻$$

Commented by bobhans last updated on 13/Jul/20

cooll....

Answered by mathmax by abdo last updated on 13/Jul/20

$${\mathrm{y}}^{{}^{″}}+{4\mathrm{y}}^{{}^{\prime }}+4\mathrm{y}=\frac{{\mathrm{e}}^{-2\mathrm{x}}}{{\mathrm{x}}^2}$$$$\left(\mathrm{he}\right)\to {\mathrm{y}}^{{}^{″}}+{4\mathrm{y}}^{{}^{\prime }}+4\mathrm{y}=0\to {\mathrm{r}}^2+4\mathrm{r}+4=0\Rightarrow {(\mathrm{r}+2)}^2=0\Rightarrow \mathrm{r}=-2$$$$\Rightarrow {\mathrm{y}}_{\mathrm{h}}=(\mathrm{ax}+\mathrm{b}){\mathrm{e}}^{-2\mathrm{x}}={\mathrm{axe}}^{-2\mathrm{x}}+{\mathrm{be}}^{-2\mathrm{x}}={\mathrm{au}}_1+{\mathrm{bu}}_2$$$$\mathrm{W}\left({\mathrm{u}}_{1,}{\mathrm{u}}_2\right)=\left|\begin{array}{c}{\mathrm{xe}}^{-2\mathrm{x}}{\mathrm{e}}^{-2\mathrm{x}}\\ (1-2\mathrm{x}){\mathrm{e}}^{-2\mathrm{x}}-{2\mathrm{e}}^{-2\mathrm{x}}\end{array}\right|=-2\mathrm{x}{\mathrm{e}}^{-2\mathrm{x}}+(2\mathrm{x}-1){\mathrm{e}}^{-2\mathrm{x}}=-{\mathrm{e}}^{-2\mathrm{x}}\ne 0$$$${\mathrm{W}}_1=\left|\begin{array}{c}\mathrm{o}{\mathrm{e}}^{-2\mathrm{x}}\\ \frac{{\mathrm{e}}^{-2\mathrm{x}}}{{\mathrm{x}}^2}-{2\mathrm{e}}^{-2\mathrm{x}}\end{array}\right|=-\frac{{\mathrm{e}}^{-4\mathrm{x}}}{{\mathrm{x}}^2}$$$${\mathrm{W}}_2=\left|\begin{array}{c}{\mathrm{xe}}^{-2\mathrm{x}}0\\ (1-2\mathrm{x}){\mathrm{e}}^{-2\mathrm{x}}\frac{{\mathrm{e}}^{-2\mathrm{x}}}{{\mathrm{x}}^2}\end{array}\right|=\frac{{\mathrm{e}}^{-4\mathrm{x}}}{\mathrm{x}}$$$${\mathrm{v}}_1=\int \frac{{\mathrm{w}}_1}{\mathrm{w}}\mathrm{dx}=\int \frac{{\mathrm{e}}^{-4\mathrm{x}}}{{\mathrm{x}}^2{\mathrm{e}}^{-2\mathrm{x}}}\mathrm{dx}=\int \frac{{\mathrm{e}}^{-2\mathrm{x}}}{{\mathrm{x}}^2}\mathrm{dx}$$$${\mathrm{v}}_2=\int \frac{{\mathrm{w}}_2}{\mathrm{w}}\mathrm{dx}=-\int \frac{{\mathrm{e}}^{-4\mathrm{x}}}{\mathrm{x}{\mathrm{e}}^{-2\mathrm{x}}}\mathrm{dx}=-\int \frac{{\mathrm{e}}^{-2\mathrm{x}}}{\mathrm{x}}\mathrm{dx}\Rightarrow$$$${\mathrm{y}}_{\mathrm{p}}={\mathrm{u}}_1{\mathrm{v}}_1+{\mathrm{u}}_2{\mathrm{v}}_2={\mathrm{xe}}^{-2\mathrm{x}}{\int }^{\mathrm{x}}\frac{{\mathrm{e}}^{-2\mathrm{t}}}{\mathrm{t}}\mathrm{dt}-{\mathrm{e}}^{-2\mathrm{x}}{\int }^{\mathrm{x}}\frac{{\mathrm{e}}^{-2\mathrm{t}}}{\mathrm{t}}\mathrm{dt}$$$$\mathrm{y}={\mathrm{y}}_{\mathrm{h}}+{\mathrm{y}}_{\mathrm{p}}$$$$$$

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