Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

Question Number 104857 by ~blr237~ last updated on 24/Jul/20

$$y^{\prime }=\frac{y-x}{y+x}y\left(0\right)=0$$

Answered by bemath last updated on 24/Jul/20

$$y=px\Rightarrow \frac{dy}{dx}=p+x\frac{dp}{dx}$$$$\iff p+x\frac{dp}{dx}=\frac{x(p-1)}{x(p+1)}$$$$x\frac{dp}{dx}=\frac{p-1}{p+1}-\frac{p(p+1)}{p+1}$$$$x\frac{dp}{dx}=\frac{-p^2-1}{p+1}\Rightarrow \frac{p+1}{p^2+1}dp=-\frac{dx}{x}$$$$\frac{1}{2}\int \frac{d(p^2+1)}{p^2+1}+\int \frac{dp}{p^2+1}=-\ln x+C$$$$\frac{1}{2}\ln (p^2+1)+{\tan }^{-1}\left(p\right)=-\ln x+C$$$$\ln \left(x\sqrt{p^2+1}\right)+{\tan }^{-1}\left(p\right)=C$$$$\ln \left(\sqrt{y^2+x^2}\right)+{\tan }^{-1}\left(\frac{y}{x}\right)=C$$

Commented by 1549442205PVT last updated on 25/Jul/20

$$\mathrm{this}\mathrm{result}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{allow}\mathrm{us}\mathrm{to}\mathrm{check}$$$$\mathrm{original}\mathrm{condition}\mathrm{easily}$$

Answered by ajfour last updated on 24/Jul/20

$$lety=tx\Rightarrow \frac{dy}{dx}=t+x\frac{dt}{dx}$$$$x\frac{dt}{dx}=\frac{t-1}{t+1}-\frac{t(t+1)}{t+1}=-\frac{(t^2+1)}{(t+1)}$$$$\Rightarrow \int \frac{t+1}{t^2+1}dt=-\int \frac{dx}{x}$$$$\frac{1}{2}\ln \mid t^2+1\mid +{\tan }^{-1}t+\ln x=c$$$$\Rightarrow \frac{1}{2}\ln \mid x^2+y^2\mid +{\tan }^{-1}\left(\frac{y}{x}\right)=c$$$$\left(noideahowtoevaluatec\right)$$

Answered by 1549442205PVT last updated on 24/Jul/20

$$\mathrm{Put}\mathrm{x}=\mathrm{yt}\Rightarrow \mathrm{dx}=\mathrm{ydt}+\mathrm{tdy}\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{y}-\mathrm{x}}{\mathrm{y}+\mathrm{x}}=\frac{1-\mathrm{t}}{\mathrm{t}+1}$$$$\frac{\mathrm{ydt}+\mathrm{tdy}}{\mathrm{dy}}=\frac{1+\mathrm{t}}{1-\mathrm{t}}\Rightarrow \frac{\mathrm{ydt}}{\mathrm{dy}}=\frac{1-{\mathrm{t}}^2}{1-\mathrm{t}}\Rightarrow \frac{\mathrm{dy}}{\mathrm{y}}=\frac{1-\mathrm{t}}{1-{\mathrm{t}}^2}\mathrm{dt}$$$$\frac{\mathrm{dy}}{\mathrm{y}}=\frac{\mathrm{dt}}{1+\mathrm{t}}\Rightarrow \ln \mid \mathrm{y}\mid =\ln \mid 1+\mathrm{t}\mid +\mathrm{lnC}(\mathrm{C}>0)$$$$\Rightarrow \mathrm{y}=\mathrm{C}(1+\mathrm{t})=\mathrm{C}(1+\frac{\mathrm{x}}{\mathrm{y}})$$$$\iff {\mathbf{y}}^2=\mathbf{C}(\mathbf{x}+\mathbf{y})\iff {\mathbf{y}}^2-\mathbf{Cy}-\mathbf{Cx}=0$$$$\mathbf{\Delta }={\mathbf{C}}^2+4\mathbf{Cx}\Rightarrow \mathbf{y}=\frac{\mathbf{C}+\sqrt{{\mathbf{C}}^2+4\mathbf{Cx}}}{2}$$$$\mathbf{or}\mathbf{y}=\frac{\mathbf{C}-\sqrt{{\mathbf{C}}^2+4\mathbf{Cx}}}{2}$$$$\mathbf{i})\mathbf{I}\mathrm{f}\mathbf{y}=\frac{\mathbf{C}+\sqrt{{\mathbf{C}}^2+4\mathbf{Cx}}}{2}\mathbf{then}$$$$\mathbf{y}\left(0\right)=0\Rightarrow 0=\frac{\mathbf{C}+\mathbf{C}}{2}\iff \mathbf{C}=0\Rightarrow \mathbf{y}\equiv 0$$$$\mathbf{contradiction}.$$$$\mathbf{ii})\mathbf{If}\mathbf{y}=\frac{\mathbf{C}-\sqrt{{\mathbf{C}}^2+4\mathbf{Cx}}}{2}\mathbf{then}$$$$\mathbf{y}\left(0\right)=0\iff 0=\frac{\mathbf{C}-\mathbf{C}}{2}\mathbf{is}\mathbf{true}\mathrm{\forall }\mathbf{C}>0$$$$\mathbf{Thus},\mathbf{y}=\frac{\mathbf{C}-\sqrt{{\mathbf{C}}^2+4\mathbf{Cx}}}{2}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com