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Question Number 12228 by tawa last updated on 16/Apr/17

$$\mathrm{Solve}\mathrm{the}\mathrm{differential}\mathrm{equation}$$$$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2\mathrm{xy}}{{\mathrm{x}}^2+{\mathrm{y}}^2}$$

Answered by mrW1 last updated on 16/Apr/17

$$u=\frac{y}{x}$$$$y=ux$$$$\frac{dy}{dx}=x\frac{du}{dx}+u$$$$\frac{dy}{dx}=\frac{2xy}{x^2+y^2}\Rightarrow$$$$x\frac{du}{dx}+u=\frac{2u}{1+u^2}$$$$x\frac{du}{dx}=\frac{2u}{1+u^2}-u=\frac{u(1-u^2)}{1+u^2}$$$$\frac{1+u^2}{u(1-u^2)}du=\frac{1}{x}dx$$$$\int \frac{1+u^2}{u(1-u^2)}du=\int \frac{1}{x}dx$$$$\frac{1}{2}\int \frac{1+u^2}{u^2(1-u^2)}{du}^2=\int \frac{1}{x}dx$$$$\frac{1}{2}\int \frac{1+w}{w(1-w)}dw=\int \frac{1}{x}dx(w=u^2)$$$$\int [\frac{1}{w}+\frac{2}{1-w}]du=2\int \frac{1}{x}dx$$$$\ln w-2\ln (1-w)=2\ln x+C_1$$$$\ln \frac{w}{x^2{(1-w)}^2}=C_1$$$$\ln \frac{u^2}{x^2{(1-u^2)}^2}=C_1$$$$2\ln \frac{u}{x(1-u^2)}=C_1$$$$\frac{u}{x(1-u^2)}=e^{\frac{C_1}{2}}=\frac{1}{C}$$$$\frac{y}{x^2(1-\frac{y^2}{x^2})}=\frac{1}{C}$$$$\frac{y}{x^2-y^2}=\frac{1}{C}$$$$y^2+Cy=x^2$$

Commented by tawa last updated on 16/Apr/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Commented by ajfour last updated on 16/Apr/17

$$errorinline6.$$

Commented by mrW1 last updated on 16/Apr/17

$$youareright.$$

Commented by tawa last updated on 16/Apr/17

$$\mathrm{please}\mathrm{sir},\mathrm{mrw}\mathrm{the}\mathrm{cubic}\mathrm{root}\mathrm{equation}\mathrm{formular}..\mathrm{i}\mathrm{need}\mathrm{the}\mathrm{image}\mathrm{again}\mathrm{sir}$$

Answered by ajfour last updated on 16/Apr/17

$$lety=txor\frac{y}{x}=t$$$$\frac{dy}{dx}=t+x\frac{dt}{dx}$$$$\Rightarrow x\frac{dt}{dx}=\frac{dy}{dx}-t$$$$orx\frac{dt}{dx}=\frac{2x\left(tx\right)}{x^2+t^2{x}^2}-t$$$$x\frac{dt}{dx}=\frac{2t}{t^2+1}-t$$$$x\frac{dt}{dx}=\frac{t(2-t^2-1)}{t^2+1}$$$$\Rightarrow \frac{dx}{x}=-\frac{(t^2+1)}{t(t-1)(t+1)}dt$$$$\int \frac{dx}{x}=-\int (\frac{A}{t}+\frac{B}{t-1}+\frac{C}{t+1})dt$$$$A=-1;B=1;C=1.$$$$so,\ln x+\ln k=\ln t-\ln (t-1)-\ln (t+1)$$$$\ln kx=\ln \frac{t}{t^2-1}$$$$kx=\frac{\left(y/x\right)}{{\left(y/x\right)}^2-1}$$$$\mathit{k}\mathit{x}=\frac{\mathit{x}\mathit{y}}{{\mathit{y}}^2-{\mathit{x}}^2}$$$$\Rightarrow \mathit{x}=0and\frac{{\mathit{y}}^2-{\mathit{x}}^2}{\mathit{y}}=\frac{1}{\mathit{k}}$$$$aresolutionstothegiven$$$$differentialeqn.$$

Commented by tawa last updated on 16/Apr/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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