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Question Number 108353 by bemath last updated on 16/Aug/20

$$\frac{△\mathcal{B}e\mathcal{M}ath△}{\dots }$$$$Generalsolutionof$$$$\frac{d^{2}y}{{dx}^2}+\frac{dy}{dx}-2y=\sin x$$

Commented by bemath last updated on 16/Aug/20

Answered by Aziztisffola last updated on 16/Aug/20

$${\mathrm{r}}^2+\mathrm{r}-2=0\Rightarrow \mathrm{r}=-2\mathrm{or}\mathrm{r}=1$$$$\Rightarrow {\mathrm{y}}_{\mathrm{h}}=\alpha {\mathrm{e}}^{-2\mathrm{x}}+\beta {\mathrm{e}}^{\mathrm{x}}$$$${\mathrm{y}}_{\mathrm{p}}={\mathrm{u}}_1{\mathrm{e}}^{-2\mathrm{x}}+{\mathrm{u}}_2{\mathrm{e}}^{\mathrm{x}}$$$$\mathrm{w}({\mathrm{u}}_1;{\mathrm{u}}_2)=\left|\begin{array}{cc}{\mathrm{e}}^{-2\mathrm{x}}& {\mathrm{e}}^{\mathrm{x}}\\ -{2\mathrm{e}}^{-2\mathrm{x}}& {\mathrm{e}}^{\mathrm{x}}\end{array}\right|={\mathrm{e}}^{-\mathrm{x}}+{2\mathrm{e}}^{-\mathrm{x}}={3\mathrm{e}}^{-\mathrm{x}}$$$${\mathrm{w}}_1=\left|\begin{array}{cc}0& {\mathrm{e}}^{\mathrm{x}}\\ \sin x& {\mathrm{e}}^{\mathrm{x}}\end{array}\right|=-{\mathrm{e}}^{\mathrm{x}}\sin \left(\mathrm{x}\right)$$$${\mathrm{w}}_2=\left|\begin{array}{cc}{\mathrm{e}}^{-2\mathrm{x}}& 0\\ -{2\mathrm{e}}^{-2\mathrm{x}}& \sin \left(\mathrm{x}\right)\end{array}\right|={\mathrm{e}}^{-2\mathrm{x}}\sin \left(\mathrm{x}\right)$$$${\mathrm{u}}_1=\int \frac{{\mathrm{w}}_1}{\mathrm{w}}\mathrm{dx}=\int \frac{-{\mathrm{e}}^{\mathrm{x}}\sin \left(\mathrm{x}\right)}{{3\mathrm{e}}^{-\mathrm{x}}}\mathrm{dx}=\frac{-1}{3}\int {\mathrm{e}}^{2\mathrm{x}}\sin \left(\mathrm{x}\right)\mathrm{dx}$$$$=\frac{-1}{3}(-{\mathrm{e}}^{2\mathrm{x}}\cos \left(\mathrm{x}\right)+2\int {\mathrm{e}}^{2\mathrm{x}}\cos \left(\mathrm{x}\right)\mathrm{dx})$$$$=-\frac{1}{15}{\mathrm{e}}^{2\mathrm{x}}(\cos \left(\mathrm{x}\right)+2\sin \left(\mathrm{x}\right))$$$${\mathrm{u}}_2=\int \frac{{\mathrm{w}}_2}{\mathrm{w}}\mathrm{dx}=\int \frac{{\mathrm{e}}^{-2\mathrm{x}}\sin \left(\mathrm{x}\right)}{{3\mathrm{e}}^{-\mathrm{x}}}\mathrm{dx}=\frac{1}{3}\int {\mathrm{e}}^{-\mathrm{x}}\sin \left(\mathrm{x}\right)\mathrm{dx}$$$$=\frac{1}{3}(-{\mathrm{e}}^{-\mathrm{x}}\cos \left(\mathrm{x}\right)+\int {\mathrm{e}}^{-\mathrm{x}}\cos \left(\mathrm{x}\right))=-\frac{1}{3}{\mathrm{e}}^{-\mathrm{x}}\cos \left(\mathrm{x}\right)$$$$\mathrm{y}={\mathrm{y}}_{\mathrm{h}}+{\mathrm{y}}_{\mathrm{p}}$$

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