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 208303 by Simurdiera last updated on 10/Jun/24

$$Resolver$$$$\frac{{\partial }^{2}u}{\partial y^2}-x^{2}u={xe}^{4y}$$

Answered by aleks041103 last updated on 12/Jun/24

$$u=X\left(x\right)Y\left(y\right)$$$$\Rightarrow XY{}^{″}-x^{2}XY=x{e}^{4y}$$$$\frac{Y{}^{″}}{Y}=\frac{x}{X}\frac{e^{4y}}{Y}+x^2$$$$leftsideisfunctionofyonly$$$$rightsidehasdependenceonx$$$$\Rightarrow \frac{Y{}^{″}}{Y}=\frac{x}{X}\frac{e^{4y}}{Y}+x^2=c=const.$$$$Y=\frac{1}{a}{e}^{4y}\Rightarrow c=16$$$$\Rightarrow 16=\frac{ax}{X}+x^2\Rightarrow X=\frac{ax}{16-x^2}$$$$\Rightarrow u=XY=\frac{{xe}^{4y}}{16-x^2}$$$$thePDEislinear$$$$\Rightarrow u=\frac{{xe}^{4y}}{16-x^2}+w$$$$where$$$$\frac{{\partial }^{2}w}{\partial y^2}-x^{2}w=0$$$$\Rightarrow w=f\left(x\right)e^{xy}+g\left(x\right)e^{-xy}$$$$ForCauchyproblem:$$$$w(x,0)=w_0\left(x\right)and\frac{\partial w}{\partial y}(x,0)=v_0\left(x\right)$$$$wecansolveforfandg.$$$$Thereforethisisthegeneralsolution.$$$$\Rightarrow u(x,y)=\frac{{xe}^{4y}}{16-x^2}+f\left(x\right)e^{xy}+g\left(x\right)e^{-xy}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com