Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 115301 by mathdave last updated on 24/Sep/20

$$if$$$${jx}^2+2kxy+{by}^2=1showthat$$$${(kx+by)}^3\frac{d^{2}y}{{dx}^2}=k^2-jb$$

Answered by 1549442205PVT last updated on 25/Sep/20

$${jx}^2+2kxy+{by}^2=1\iff {\mathrm{by}}^2+2\mathrm{kxy}+{\mathrm{jx}}^2-1=0$$$${\mathrm{\Delta }}^{\prime }={\mathrm{k}}^2{\mathrm{x}}^2-{\mathrm{jbx}}^2+\mathrm{b}$$$$\mathrm{y}=\frac{-\mathrm{kx}\pm \sqrt{{\mathrm{k}}^2{\mathrm{x}}^2-{\mathrm{jbx}}^2+\mathrm{b}}}{\mathrm{b}}\left(1\right)$$$${\mathrm{y}}^{\prime }=(-\mathrm{k}\pm \frac{({\mathrm{k}}^2-\mathrm{jb})\mathrm{x}}{\sqrt{{\mathrm{k}}^2{\mathrm{x}}^2-{\mathrm{jbx}}^2+\mathrm{b}}})/\mathrm{b}$$$${\mathrm{y}}^{″}=\pm \left(\frac{({\mathrm{k}}^2-\mathrm{jb})\sqrt{{\mathrm{k}}^2{\mathrm{x}}^2-{\mathrm{jbx}}^2+\mathrm{b}}-({\mathrm{k}}^2-\mathrm{jb})\mathrm{x}.\frac{({\mathrm{k}}^2-\mathrm{jb})\mathrm{x}}{\sqrt{{\mathrm{k}}^2{\mathrm{x}}^2-{\mathrm{jbx}}^2+\mathrm{b}}}}{{\mathrm{k}}^2{\mathrm{x}}^2-{\mathrm{jbx}}^2+\mathrm{b}}\right)/\mathrm{b}$$$$=\pm \left(\frac{({\mathrm{k}}^2-\mathrm{jb})({\mathrm{k}}^2{\mathrm{x}}^2-{\mathrm{jbx}}^2+\mathrm{b})-{({\mathrm{k}}^2-\mathrm{jb})}^2{\mathrm{x}}^2}{\mathrm{b}({\mathrm{k}}^2{\mathrm{x}}^2-{\mathrm{jbx}}^2+\mathrm{b})\sqrt{{\mathrm{k}}^2{\mathrm{x}}^2-{\mathrm{jbx}}^2+\mathrm{b}}}\right)$$$$=\pm \left(\frac{({\mathrm{k}}^2-\mathrm{jb})}{({\mathrm{k}}^2{\mathrm{x}}^2-{\mathrm{jbx}}^2+\mathrm{b})\sqrt{{\mathrm{k}}^2{\mathrm{x}}^2-{\mathrm{jbx}}^2+\mathrm{b}}}\right)\left(2\right)$$$$\mathrm{From}\left(1\right)\mathrm{we}\mathrm{have}$$$$\mathrm{kx}+\mathrm{by}=\pm \sqrt{{\mathrm{k}}^2{\mathrm{x}}^2-{\mathrm{jbx}}^2+\mathrm{b}}$$$$\Rightarrow {(\mathrm{kx}+\mathrm{by})}^3=({\mathrm{k}}^2{\mathrm{x}}^2-{\mathrm{jbx}}^2+\mathrm{b})\sqrt{{\mathrm{k}}^2{\mathrm{x}}^2-{\mathrm{jbx}}^2+\mathrm{b}}\left(3\right)$$$$\mathrm{Replace}\left(3\right)\mathrm{into}\left(2\right)\mathrm{we}\mathrm{get}$$$${(\mathbf{kx}+\mathbf{by})}^3{\mathbf{y}}^{″}={\mathbf{k}}^2-\mathbf{jb}(\mathbf{Q}.\mathbf{E}.\mathbf{D})$$

Answered by Dwaipayan Shikari last updated on 25/Sep/20

$${\mathrm{jx}}^2+2\mathrm{kxy}+{\mathrm{by}}^2=1$$$$2\mathrm{jx}+2\mathrm{ky}+2\mathrm{kx}\frac{\mathrm{dy}}{\mathrm{dx}}+2\mathrm{by}\frac{\mathrm{dy}}{\mathrm{dx}}=0\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{\mathrm{jx}+\mathrm{ky}}{\mathrm{kx}+\mathrm{by}}$$$$2\mathrm{j}+2\mathrm{k}.\frac{\mathrm{dy}}{\mathrm{dx}}+2\mathrm{kx}\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}+2\mathrm{k}.\frac{\mathrm{dy}}{\mathrm{dx}}+2\mathrm{by}\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}+2\mathrm{b}{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2=0$$$$\mathrm{j}+2\mathrm{k}\frac{\mathrm{dy}}{\mathrm{dx}}+\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}(\mathrm{kx}+\mathrm{by})+\mathrm{b}{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2=0$$$$\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}(\mathrm{kx}+\mathrm{by})=-\mathrm{b}{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2+2\mathrm{k}.\frac{\mathrm{jx}+\mathrm{ky}}{\mathrm{kx}+\mathrm{by}}-\mathrm{j}$$$$\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}(\mathrm{kx}+\mathrm{by})=-\mathrm{b}{\left(\frac{\mathrm{jx}+\mathrm{ky}}{\mathrm{kx}+\mathrm{by}}\right)}^2+2\mathrm{k}\frac{\mathrm{jx}+\mathrm{ky}}{\mathrm{kx}+\mathrm{by}}-\mathrm{j}$$$$\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}{(\mathrm{kx}+\mathrm{by})}^3=-\mathrm{b}{(\mathrm{jx}+\mathrm{ky})}^2+2\mathrm{k}(\mathrm{jx}+\mathrm{ky})(\mathrm{kx}+\mathrm{by})-\mathrm{j}{(\mathrm{kx}+\mathrm{by})}^2$$$$\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}{(\mathrm{kx}+\mathrm{by})}^3={\mathrm{k}}^2-\mathrm{jb}\left(\mathrm{After}\mathrm{simplifying}\right)$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com