Question and Answers Forum

All Questions      Topic List

Differential Equation Questions

Previous in All Question      Next in All Question      

Previous in Differential Equation      Next in Differential Equation      

Question Number 94328 by i jagooll last updated on 18/May/20

$${\mathrm{y}}^{\prime }+\mathrm{xy}=\mathrm{x}$$

Answered by i jagooll last updated on 18/May/20

Commented by i jagooll last updated on 18/May/20

$$\mathrm{it}\mathrm{correct}?$$

Answered by abdomathmax last updated on 18/May/20

$$\left(\mathrm{he}\right)\to {\mathrm{y}}^{{}^{\prime }}+\mathrm{xy}=0\to \frac{{\mathrm{y}}^{{}^{\prime }}}{\mathrm{y}}=-\mathrm{x}\Rightarrow \ln \mid \mathrm{y}\mid =-\frac{{\mathrm{x}}^2}{2}+\lambda \Rightarrow$$$$\mathrm{y}=\mathrm{k}{\mathrm{e}}^{-\frac{{\mathrm{x}}^2}{2}}\mathrm{let}\mathrm{use}\mathrm{mvc}\mathrm{method}\left(\mathrm{lagrange}\right)$$$${\mathrm{y}}^{{}^{\prime }}={\mathrm{k}}^{{}^{\prime }}{\mathrm{e}}^{-\frac{{\mathrm{x}}^2}{2}}-\mathrm{xk}{\mathrm{e}}^{-\frac{{\mathrm{x}}^2}{2}}$$$$\left(\mathrm{e}\right)\Rightarrow {\mathrm{k}}^{{}^{\prime }}{\mathrm{e}}^{-\frac{{\mathrm{x}}^2}{2}}-\mathrm{kx}{\mathrm{e}}^{-\frac{{\mathrm{x}}^2}{2}}+\mathrm{kx}{\mathrm{e}}^{-\frac{{\mathrm{x}}^2}{2}}=\mathrm{x}\Rightarrow$$$${\mathrm{k}}^{{}^{\prime }}=\mathrm{x}{\mathrm{e}}^{\frac{{\mathrm{x}}^2}{2}}\Rightarrow \mathrm{k}=\int \mathrm{x}{\mathrm{e}}^{\frac{{\mathrm{x}}^2}{2}}+\mathrm{c}={\mathrm{e}}^{\frac{{\mathrm{x}}^2}{2}}+\mathrm{c}\Rightarrow$$$$\mathrm{y}\left(\mathrm{x}\right)=({\mathrm{e}}^{\frac{{\mathrm{x}}^2}{2}}+\mathrm{c}){\mathrm{e}}^{-\frac{{\mathrm{x}}^2}{2}}\Rightarrow \mathrm{y}\left(\mathrm{x}\right)=1+\mathrm{c}{\mathrm{e}}^{-\frac{{\mathrm{x}}^2}{2}}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com