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Question Number 92702 by i jagooll last updated on 08/May/20

$$y^{″}+2y^{\prime }-3y=e^x+e^{2x}$$

Answered by niroj last updated on 08/May/20

$${\mathrm{y}}^{{}^{″}}+{2\mathrm{y}}^{{}^{\prime }}-3\mathrm{y}={\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{2\mathrm{x}}$$$$\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}+2\frac{\mathrm{dy}}{\mathrm{dx}}-3\mathrm{y}={\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{2\mathrm{x}}$$$$({\mathrm{D}}^2+2\mathrm{D}-3)\mathrm{y}={\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{2\mathrm{x}}$$$$\mathrm{A}.\mathrm{E}.{\mathrm{m}}^2+2\mathrm{m}-3=0$$$${\mathrm{m}}^2+3\mathrm{x}-\mathrm{m}-3=0$$$$\mathrm{m}(\mathrm{m}+3)-1(\mathrm{m}+3)=0$$$$(\mathrm{m}+3)(\mathrm{m}-1)=0$$$$\mathrm{m}=1,-3$$$$\mathrm{CF}={\mathrm{C}}_1{\mathrm{e}}^{\mathrm{x}}+{\mathrm{C}}_2{\mathrm{e}}^{-3\mathrm{x}}$$$$\mathrm{PI}=\frac{1}{{\mathrm{D}}^2+2\mathrm{D}-3}({\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{2\mathrm{x}})$$$$=\frac{{\mathrm{e}}^{\mathrm{x}}}{{\mathrm{D}}^2+2\mathrm{D}-3}+\frac{{\mathrm{e}}^{2\mathrm{x}}}{{\mathrm{D}}^2+2\mathrm{D}-3}$$$$=\frac{{\mathrm{xe}}^{\mathrm{x}}}{2\mathrm{D}+2}+\frac{{\mathrm{e}}^{2\mathrm{x}}}{4+4-3}$$$$=\frac{{\mathrm{xe}}^{\mathrm{x}}}{2\times 1+2}+\frac{{\mathrm{e}}^{2\mathrm{x}}}{5}$$$$=\frac{{\mathrm{xe}}^{\mathrm{x}}}{4}+\frac{{\mathrm{e}}^{2\mathrm{x}}}{5}$$$$\mathrm{y}=\mathrm{CF}+\mathrm{PI}$$$$\mathrm{y}={\mathrm{C}}_1{\mathrm{e}}^{\mathrm{x}}+{\mathrm{C}}_1{\mathrm{e}}^{-3\mathrm{x}}+\frac{{\mathrm{xe}}^{\mathrm{x}}}{4}+\frac{{\mathrm{e}}^{2\mathrm{x}}}{5}//.$$

Answered by Rio Michael last updated on 08/May/20

$$\mathrm{the}\mathrm{A}-\mathrm{level}\mathrm{thought}\mathrm{method}.$$$$\mathrm{auxillary}\mathrm{equation}:m^2+2m-3=0\Rightarrow (m+3)(m-1)$$$$\iff m=-3\mathrm{or}m=1$$$$y_{c.f}=A{e}^x+{Be}^{-3x}$$$$\mathrm{now}{y}_{p.i}=\lambda {xe}^x+\mu e^{2x}$$$$y^{\prime }=\lambda {xe}^x+\lambda e^x+2\mu e^{2x}$$$$y^{″}=\lambda {xe}^x+2\lambda e^x+4\mu e^{2x}$$$$\Rightarrow \lambda {xe}^x+2\lambda e^x+4\mu e^{2x}+2\lambda {xe}^x+2\lambda e^x+4\mu e^{2x}-3\lambda {xe}^x-3\mu e^{2x}=e^x+e^{2x}$$$$\Rightarrow (4\lambda -1)e^x+(5\mu -1)e^{2x}=0$$$$4\lambda -1=0\mathrm{and}5\mu -1=0$$$$\iff \lambda =\frac{1}{4}\mathrm{and}\mu =\frac{1}{5}$$$$y_{p.i}=\frac{{xe}^x}{4}+\frac{e^{2x}}{5}$$$$y=y_{cf}+y_{p.i}$$$$y={Ae}^x+{Be}^{-x}+\frac{{xe}^x}{4}+\frac{e^{2x}}{5}$$

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