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Question Number 91024 by john santu last updated on 27/Apr/20

$$x^2{y}^{″}+3{xy}^{\prime }+2y=4x^2$$

Commented by jagoll last updated on 27/Apr/20

$$icantry$$$$letx=e^z$$$$auxillaryequation$$$$r^2+2r+2=0$$$$r=-1\pm i$$$$y_c=e^{-z}\{C_1\cos \left(z\right)+C_2\sin \left(z\right)\}$$$$(D^2+2D+2)y=4e^{2z}$$$$particularsolution$$$$y_p=4\left(\frac{1}{D^2+2D+2}\right)e^{2z}$$$$y_p=\frac{4}{10}{e}^{2z}=\frac{2}{5}{x}^2$$$$completesolution$$$$y=e^{-z}\{C_1\cos \left(z\right)+C_2\sin \left(z\right)\}+\frac{2}{5}{x}^2$$$$y=\frac{1}{x}\{C_1\cos \left(\ln x\right)+C_2\sin \left(\ln x\right)\}+\frac{2x^2}{5}$$

Commented by niroj last updated on 27/Apr/20

$${\mathrm{x}}^2{\mathrm{y}}^{{}^{″}}+{3\mathrm{xy}}^{{}^{\prime }}+2\mathrm{y}={4\mathrm{x}}^2$$$${\mathrm{x}}^2\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}+3\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}+2\mathrm{y}={4\mathrm{x}}^2$$$$\mathrm{Put},{\mathrm{x}}^2\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=({\mathrm{D}}^2-\mathrm{D})\mathrm{y},\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{Dy}$$$$\mathrm{and}\mathrm{x}={\mathrm{e}}^{\mathrm{z}}\mathrm{or}\mathrm{z}=\log \mathrm{x}$$$$({\mathrm{D}}^2-\mathrm{D}+3\mathrm{D}+2)\mathrm{y}=4{\mathrm{e}}^{2\mathrm{z}}$$$$({\mathrm{D}}^2+2\mathrm{D}+2)\mathrm{y}={4\mathrm{e}}^{2\mathrm{z}}$$$$\mathrm{Auxiliary}\mathrm{Equation},$$$${\mathrm{m}}^2+2\mathrm{m}+2=0$$$$\mathrm{m}=\frac{-2\stackrel{-}{+}\sqrt{4-8}}{2}=\frac{-2\stackrel{-}{+}\sqrt{-4}}{2}$$$$=\frac{-2\stackrel{-}{+}\sqrt{-4}}{2}=\frac{-2\stackrel{-}{+}\sqrt{4\mathrm{i}}}{2}=\frac{-2\stackrel{-}{+}2\mathrm{i}}{2}$$$$\mathrm{m}=-1\stackrel{-}{+}1\mathrm{i}$$$$\mathrm{CF}={\mathrm{e}}^{-\mathrm{z}}({\mathrm{C}}_1\cos \mathrm{z}+{\mathrm{C}}_2\mathrm{sinz})$$$$\mathrm{Now}\mathrm{again},$$$$\mathrm{PI}=\frac{{4\mathrm{e}}^{2\mathrm{z}}}{{\mathrm{D}}^2+2\mathrm{D}+2}=\frac{{4\mathrm{e}}^{2\mathrm{z}}}{4+4+2}=\frac{{4\mathrm{e}}^{2\mathrm{z}}}{10}$$$$\mathrm{PI}=\frac{{2\mathrm{e}}^{2\mathrm{z}}}{5}$$$$\mathrm{Complete}\mathrm{Solution},$$$$\mathrm{y}=\mathrm{CF}+\mathrm{PI}$$$$={\mathrm{e}}^{-\mathrm{z}}({\mathrm{C}}_1\cos \mathrm{z}+{\mathrm{C}}_2\mathrm{sinz})+\frac{{2\mathrm{e}}^{2\mathrm{z}}}{5}$$$$\mathrm{put}\mathrm{z}=\log \mathrm{x}$$$$\mathrm{y}={\mathrm{e}}^{-\log \mathrm{x}}({\mathrm{C}}_1\cos \log \mathrm{x}+{\mathrm{C}}_2\sin \log \mathrm{x})+\frac{{2\mathrm{e}}^{\log {\mathrm{x}}^2}}{5}$$$$\mathrm{y}=\frac{1}{\mathrm{x}}({\mathrm{C}}_1\cos \log \mathrm{x}+{\mathrm{C}}_2\sin \log \mathrm{x})+\frac{2}{5}{\mathrm{x}}^2//.$$$$$$

Commented by jagoll last updated on 27/Apr/20

$$whyPI=\frac{4e^{2z}}{14}=\frac{2e^z}{7}how?$$

Commented by MWSuSon last updated on 27/Apr/20

$$checkyour$$$$ithinkitsamistake$$

Commented by niroj last updated on 27/Apr/20

$$yessdearitsnowedited.$$

Answered by MWSuSon last updated on 27/Apr/20

$$letx=e^z$$$$x^2{y}^{″}=D(D-1)y$$$${xy}^{\prime }=Dy$$$$D(D-1)y+3Dy+2y=4e^{2z}$$$$(D^2+2D+2)y=4e^{2z}$$$$Auxillaryeqution$$$$m^2+2m+2=0$$$$m=-1\pm i$$$$y_c=e^{-z}(C_1\cos \left(z\right)+C_2\sin \left(z\right))$$$$y_p=4\frac{1}{D^2+2D+2}{e}^{2z}$$$$=4\frac{1}{4+4+2}{e}^{2z}$$$$=\frac{4e^{2z}}{10}$$$$=\frac{2e^{2z}}{5}$$$$y=e^{-z}(C_1\cos \left(z\right)+C_2\sin \left(z\right))+\frac{2}{5}{e}^{2z}$$$$butz={log}_{e}x$$$$y=\frac{1}{x}[C_1\cos \left({log}_{e}x\right)+C_2\sin \left({log}_{e}x\right)]+\frac{2}{5}{x}^2$$

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