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 89205 by jagoll last updated on 16/Apr/20

$$2\frac{dy}{dx}=\frac{y}{x}+{\left(\frac{y}{x}\right)}^2$$

Commented by john santu last updated on 16/Apr/20

$$y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$$$$2v+2x\frac{dv}{dx}=v+v^2$$$$2x\frac{dv}{dx}=v^2-v$$$$\frac{dv}{v(v-1)}=\frac{dx}{2x}$$$$\int (\frac{1}{2(v-1)}-\frac{1}{2v})dv=\int \frac{dx}{2x}$$$$\frac{1}{2}\ln \mid \frac{v-1}{v}\mid =\frac{1}{2}\ln \mid x\mid +c$$$$\ln \mid 1-\frac{x}{y}\mid =\ln \mid x\mid +C$$$$\pm (1-\frac{x}{y})={ke}^x$$$$1-\frac{x}{y}=\pm {ke}^x$$$$\frac{x}{y}=1\pm {ke}^x\Rightarrow y=\frac{x}{1\pm {ke}^x}$$

Answered by 242242864 last updated on 16/Apr/20

$$\mathrm{let}\mathrm{u}=\frac{y}{x}$$$$\Rightarrow y=ux$$$$\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=\mathit{u}+\mathit{x}\frac{\mathit{d}\mathit{u}}{\mathit{d}\mathit{x}}$$$$2[\mathit{u}+\mathit{x}\frac{\mathit{d}\mathit{u}}{\mathit{d}\mathit{x}}]=\mathit{u}+{\mathit{u}}^2$$$$2\mathit{u}+2\mathit{x}\frac{\mathit{d}\mathit{u}}{\mathit{d}\mathit{c}}=\mathit{u}+{\mathit{u}}^2$$$$2\mathit{x}\frac{\mathit{d}\mathit{u}}{\mathit{d}\mathit{x}}={\mathit{u}}^2-\mathit{u}$$$$2\mathit{x}\mathit{d}\mathit{u}=({\mathit{u}}^2-\mathit{u})\mathit{d}\mathit{x}$$$$\frac{2\mathit{d}\mathit{u}}{({\mathit{u}}^2-\mathit{u})}=\frac{\mathit{d}\mathit{x}}{\mathit{x}}$$$$2\int \frac{\mathit{d}\mathit{u}}{({\mathit{u}}^2-\mathit{u})}=\int \frac{\mathit{d}\mathit{x}}{\mathit{x}}$$$$\mathit{c}\mathit{o}\mathit{n}\mathit{s}\mathit{i}\mathit{d}\mathit{e}\mathit{r}\frac{1}{({\mathit{u}}^2-\mathit{u})}\mathit{t}\mathit{o}\mathit{b}\mathit{e}\mathit{c}\mathit{h}\mathit{a}\mathit{n}\mathit{g}\mathit{e}\mathit{t}\mathit{o}\mathit{p}\mathit{a}\mathit{r}\mathit{t}\mathit{i}\mathit{a}\mathit{l}\mathit{f}\mathit{r}\mathit{a}\mathit{c}\mathit{t}\mathit{i}\mathit{o}\mathit{n},$$$$\frac{1}{({\mathit{u}}^2-\mathit{u})}=\frac{1}{(\mathit{u}+1)(\mathit{u}-1)}$$$$\Rightarrow \frac{1}{(\mathit{u}+1)(\mathit{u}-1)}=\frac{A}{(\mathit{u}+1)}+\frac{B}{(\mathit{u}-1)}$$$$1=A(\mathit{u}-1)+B(\mathit{u}+1)$$$$1=A\mathit{u}-A+B\mathit{u}+B$$$$1=\mathit{u}(A+B)+B-A$$$$B-A=1.........\left(i\right)$$$$A+B=0..........\left(ii\right)$$$$\Rightarrow A=-\frac{1}{2},B=\frac{1}{2}$$$$\Rightarrow \int \frac{1}{({\mathit{u}}^2-1)}\mathit{d}\mathit{u}=\frac{1}{2}\int \frac{\mathit{d}\mathit{u}}{(\mathit{u}-1)}-\frac{1}{2}\int \frac{\mathit{d}\mathit{u}}{(\mathit{u}+1)}=\int \frac{\mathit{d}\mathit{x}}{\mathit{x}}$$$$\frac{1}{2}\mathit{l}\mathit{n}(\mathit{u}-1)-\frac{1}{2}\mathit{l}\mathit{n}(\mathit{u}+1)=\mathit{l}\mathit{n}\mathit{x}+C$$$$\frac{1}{2}\mathit{l}\mathit{n}\left[\frac{\mathit{u}-1}{\mathit{u}+1}\right]=\mathit{l}\mathit{n}\mathit{x}+C$$$$\frac{1}{2}\mathit{l}\mathit{n}\left[\frac{\mathit{y}-\mathit{x}}{\mathit{y}+\mathit{x}}\right]=\mathit{l}\mathit{n}\mathit{x}+C$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com