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Question Number 94956 by i jagooll last updated on 22/May/20

Commented by niroj last updated on 22/May/20

$$\mathrm{both}\mathrm{of}\mathrm{you}\mathrm{are}\mathrm{right}\mathrm{its}\mathrm{note}:$$

Commented by i jagooll last updated on 22/May/20

$$\mathrm{edit}x\frac{dy}{dx}+3x(3x+1)y=e^{-3x}$$

Commented by niroj last updated on 22/May/20

$$\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}+3\mathrm{x}(3\mathrm{x}+1)\mathrm{y}={\mathrm{e}}^{-3\mathrm{x}}$$$$\frac{\mathrm{dy}}{\mathrm{dx}}+\frac{3\mathrm{x}(3\mathrm{x}+1)}{\mathrm{x}}\mathrm{y}=\frac{{\mathrm{e}}^{-3\mathrm{x}}}{\mathrm{x}}$$$$\frac{\mathrm{dy}}{\mathrm{dx}}+3(3\mathrm{x}+1)\mathrm{y}=\frac{{\mathrm{e}}^{-3\mathrm{x}}}{\mathrm{x}}$$$$\mathrm{P}=3(3\mathrm{x}+1),\mathrm{Q}=\frac{{\mathrm{e}}^{-3\mathrm{x}}}{\mathrm{x}}$$$$\mathrm{IF}={\mathrm{e}}^{\int \mathrm{Pdx}}={\mathrm{e}}^{\int 3(3\mathrm{x}+1)\mathrm{dx}}$$$$={\mathrm{e}}^{9\int \mathrm{xdx}+3\int \mathrm{dx}}$$$$={\mathrm{e}}^{9\times \frac{{\mathrm{x}}^2}{2}+3\mathrm{x}}$$$$={\mathrm{e}}^{\frac{9}{2}{\mathrm{x}}^2}.{\mathrm{e}}^{3\mathrm{x}}$$$$\mathrm{y}.{\mathrm{e}}^{\frac{9}{2}{\mathrm{x}}^2}.{\mathrm{e}}^{3\mathrm{x}}=\int {\mathrm{e}}^{\frac{9}{2}{\mathrm{x}}^2}.{\mathrm{e}}^{3\mathrm{x}}.\frac{{\mathrm{e}}^{-3\mathrm{x}}}{\mathrm{x}}\mathrm{dx}+\mathrm{C}$$$$\mathrm{y}{\mathrm{e}}^{\frac{9}{2}{\mathrm{x}}^2+3\mathrm{x}}=\int \frac{1}{\mathrm{x}}{\mathrm{e}}^{\frac{9}{2}{\mathrm{x}}^2}\mathrm{dx}+\mathrm{C}$$$$$$$$\mathrm{y}=\frac{1}{{\mathrm{e}}^{\frac{9}{2}{\mathrm{x}}^2+3\mathrm{x}}}[\int \frac{1}{\mathrm{x}}{\mathrm{e}}^{\frac{9}{2}{\mathrm{x}}^2}\mathrm{dx}+\mathrm{C}]..$$$$\mathrm{now}\mathrm{you}\mathrm{can}\mathrm{runway}..\mathrm{free}..\mathrm{ly}$$

Commented by bobhans last updated on 22/May/20

hahaha��������

Commented by bobhans last updated on 22/May/20

$$\mathrm{i}\mathrm{think}\mathrm{in}{9}^{\mathrm{th}}\mathrm{line}\mathrm{is}\mathrm{wrong}.\mathrm{it}$$$$\mathrm{should}\mathrm{be}\int \frac{{\mathrm{e}}^{\frac{9}{2}{x}^2}}{x}dx$$

Commented by i jagooll last updated on 22/May/20

$$\mathrm{how}\int \frac{{\mathrm{e}}^{\frac{9}{2}{x}^2}dx}{x}\Rightarrow \int {xe}^{\frac{9}{2}{x}^2}dxsir$$

Commented by peter frank last updated on 22/May/20

$$\mathrm{thank}\mathrm{you}$$

Answered by mathmax by abdo last updated on 22/May/20

$${\mathrm{xy}}^{{}^{\prime }}+3\mathrm{x}(3\mathrm{x}+1)\mathrm{y}={\mathrm{e}}^{-3\mathrm{x}}$$$$\left(\mathrm{he}\right)\to {\mathrm{xy}}^{{}^{\prime }}=-3\mathrm{x}(3\mathrm{x}+1)\mathrm{y}\Rightarrow \frac{{\mathrm{y}}^{\prime }}{\mathrm{y}}=-3(3\mathrm{x}+1)\Rightarrow$$$$\ln \mid \mathrm{y}\mid =-3\int (3\mathrm{x}+1)\mathrm{dx}=-3(\frac{3}{2}{\mathrm{x}}^2+\mathrm{x})=-\frac{9}{2}{\mathrm{x}}^2-3\mathrm{x}+\mathrm{c}\Rightarrow$$$$\mathrm{y}=\mathrm{k}{\mathrm{e}}^{-\frac{9}{2}{\mathrm{x}}^2-3\mathrm{x}}\mathrm{mvc}\mathrm{method}\Rightarrow {\mathrm{y}}^{{}^{\prime }}=(-9\mathrm{x}-3)\mathrm{k}{\mathrm{e}}^{-\frac{9}{2}{\mathrm{x}}^2-3\mathrm{x}}$$$$+{\mathrm{k}}^{{}^{\prime }}{\mathrm{e}}^{-\frac{9}{2}{\mathrm{x}}^2-3\mathrm{x}}$$$$\left(\mathrm{e}\right)\Rightarrow (-{9\mathrm{x}}^2-3\mathrm{x}){\mathrm{e}}^{-\frac{9}{2}{\mathrm{x}}^2-3\mathrm{x}}+{\mathrm{xk}}^{{}^{\prime }}{\mathrm{e}}^{-\frac{9}{2}{\mathrm{x}}^2-3\mathrm{x}}+({9\mathrm{x}}^2+3\mathrm{x})\mathrm{k}{\mathrm{e}}^{-\frac{9}{2}{\mathrm{x}}^2-3\mathrm{x}}$$$$={\mathrm{e}}^{-3\mathrm{x}}\Rightarrow {\mathrm{xk}}^{{}^{\prime }}{\mathrm{e}}^{-\frac{9}{2}{\mathrm{x}}^2-3\mathrm{x}}={\mathrm{e}}^{-3\mathrm{x}}\Rightarrow {\mathrm{k}}^{\prime }=\frac{1}{\mathrm{x}}{\mathrm{e}}^{\frac{9}{2}{\mathrm{x}}^2}\Rightarrow$$$$\mathrm{k}\left(\mathrm{x}\right)={\int }_{.}^{\mathrm{x}}\frac{1}{\mathrm{t}}{\mathrm{e}}^{\frac{9}{2}{\mathrm{t}}^2}\mathrm{dt}+\mathrm{c}\Rightarrow \mathrm{y}\left(\mathrm{x}\right)=({\int }_{.}^{\mathrm{x}}\frac{1}{\mathrm{t}}{\mathrm{e}}^{\frac{9}{2}{\mathrm{t}}^2}\mathrm{dt}+\mathrm{c}){\mathrm{e}}^{-\frac{9}{2}{\mathrm{x}}^2-3\mathrm{x}}$$

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