(A): Given that the differential equation is \((x+y)^{2}\left(x \frac{d y}{d x}+y\right)=x y\left(1+\frac{d y}{d x}\right)\) \((x y)^{-1}\left(x \frac{d y}{d x}+y\right)=(x+y)^{-2}\left(1+\frac{d y}{d x}\right)\) Integrating on both sides we get - \(\int(x y)^{-1}\left(x \frac{d y}{d x}+y\right)=\int(x+y)^{-2}\left(1+\frac{d y}{d x}\right)\) We know that integral \(\int \frac{\mathrm{f}^{\prime}(\mathrm{x})}{\mathrm{f}(\mathrm{x})} \mathrm{dx}=\log (\mathrm{f}(\mathrm{x}))+\mathrm{C}\) and Put, \(\quad(\mathrm{x}+\mathrm{y})=\mathrm{t} \Rightarrow\left(1+\frac{\mathrm{dy}}{\mathrm{dx}}\right)=\mathrm{dt}\) \(\log (\mathrm{xy})=\int \mathrm{t}^{-2} \mathrm{dt} \Rightarrow \log (\mathrm{xy})=\frac{\mathrm{t}^{-1}}{-1}+\mathrm{c}\) \(\log (x y)=\frac{(x+y)^{-1}}{-1}+C\) \(\log (x y)=\frac{-1}{x+y}+C\)
COMEDK-2014
Differential Equation
87206
The differential equation of \(y=\mathbf{a e}^{\mathbf{b x + c}}\) is
(C) : Given that the differential equation \(y=a e^{b x+c} \tag{i}\) Differentiating equation (i) w.r.t. \(x\), we get - \(y_{1}=a\left(e^{b x+c}\right)(b)=b\left(a e^{b x+c}\right)=b y\) \(b=\frac{y_{1}}{y} \tag{ii}\) Again differentiating equation (ii) w.r.t. \(\mathrm{x}\), we get - \(0=\frac{y \cdot y_{2}-y_{1} \cdot y_{1}}{y^{2}} \Rightarrow y \cdot y_{2}-y_{1} y_{1}=0\) \(y_{1}^{2}=y \cdot y_{2}\)
(A): Given that the differential equation is \((x+y)^{2}\left(x \frac{d y}{d x}+y\right)=x y\left(1+\frac{d y}{d x}\right)\) \((x y)^{-1}\left(x \frac{d y}{d x}+y\right)=(x+y)^{-2}\left(1+\frac{d y}{d x}\right)\) Integrating on both sides we get - \(\int(x y)^{-1}\left(x \frac{d y}{d x}+y\right)=\int(x+y)^{-2}\left(1+\frac{d y}{d x}\right)\) We know that integral \(\int \frac{\mathrm{f}^{\prime}(\mathrm{x})}{\mathrm{f}(\mathrm{x})} \mathrm{dx}=\log (\mathrm{f}(\mathrm{x}))+\mathrm{C}\) and Put, \(\quad(\mathrm{x}+\mathrm{y})=\mathrm{t} \Rightarrow\left(1+\frac{\mathrm{dy}}{\mathrm{dx}}\right)=\mathrm{dt}\) \(\log (\mathrm{xy})=\int \mathrm{t}^{-2} \mathrm{dt} \Rightarrow \log (\mathrm{xy})=\frac{\mathrm{t}^{-1}}{-1}+\mathrm{c}\) \(\log (x y)=\frac{(x+y)^{-1}}{-1}+C\) \(\log (x y)=\frac{-1}{x+y}+C\)
COMEDK-2014
Differential Equation
87206
The differential equation of \(y=\mathbf{a e}^{\mathbf{b x + c}}\) is
(C) : Given that the differential equation \(y=a e^{b x+c} \tag{i}\) Differentiating equation (i) w.r.t. \(x\), we get - \(y_{1}=a\left(e^{b x+c}\right)(b)=b\left(a e^{b x+c}\right)=b y\) \(b=\frac{y_{1}}{y} \tag{ii}\) Again differentiating equation (ii) w.r.t. \(\mathrm{x}\), we get - \(0=\frac{y \cdot y_{2}-y_{1} \cdot y_{1}}{y^{2}} \Rightarrow y \cdot y_{2}-y_{1} y_{1}=0\) \(y_{1}^{2}=y \cdot y_{2}\)
(A): Given that the differential equation is \((x+y)^{2}\left(x \frac{d y}{d x}+y\right)=x y\left(1+\frac{d y}{d x}\right)\) \((x y)^{-1}\left(x \frac{d y}{d x}+y\right)=(x+y)^{-2}\left(1+\frac{d y}{d x}\right)\) Integrating on both sides we get - \(\int(x y)^{-1}\left(x \frac{d y}{d x}+y\right)=\int(x+y)^{-2}\left(1+\frac{d y}{d x}\right)\) We know that integral \(\int \frac{\mathrm{f}^{\prime}(\mathrm{x})}{\mathrm{f}(\mathrm{x})} \mathrm{dx}=\log (\mathrm{f}(\mathrm{x}))+\mathrm{C}\) and Put, \(\quad(\mathrm{x}+\mathrm{y})=\mathrm{t} \Rightarrow\left(1+\frac{\mathrm{dy}}{\mathrm{dx}}\right)=\mathrm{dt}\) \(\log (\mathrm{xy})=\int \mathrm{t}^{-2} \mathrm{dt} \Rightarrow \log (\mathrm{xy})=\frac{\mathrm{t}^{-1}}{-1}+\mathrm{c}\) \(\log (x y)=\frac{(x+y)^{-1}}{-1}+C\) \(\log (x y)=\frac{-1}{x+y}+C\)
COMEDK-2014
Differential Equation
87206
The differential equation of \(y=\mathbf{a e}^{\mathbf{b x + c}}\) is
(C) : Given that the differential equation \(y=a e^{b x+c} \tag{i}\) Differentiating equation (i) w.r.t. \(x\), we get - \(y_{1}=a\left(e^{b x+c}\right)(b)=b\left(a e^{b x+c}\right)=b y\) \(b=\frac{y_{1}}{y} \tag{ii}\) Again differentiating equation (ii) w.r.t. \(\mathrm{x}\), we get - \(0=\frac{y \cdot y_{2}-y_{1} \cdot y_{1}}{y^{2}} \Rightarrow y \cdot y_{2}-y_{1} y_{1}=0\) \(y_{1}^{2}=y \cdot y_{2}\)
(A): Given that the differential equation is \((x+y)^{2}\left(x \frac{d y}{d x}+y\right)=x y\left(1+\frac{d y}{d x}\right)\) \((x y)^{-1}\left(x \frac{d y}{d x}+y\right)=(x+y)^{-2}\left(1+\frac{d y}{d x}\right)\) Integrating on both sides we get - \(\int(x y)^{-1}\left(x \frac{d y}{d x}+y\right)=\int(x+y)^{-2}\left(1+\frac{d y}{d x}\right)\) We know that integral \(\int \frac{\mathrm{f}^{\prime}(\mathrm{x})}{\mathrm{f}(\mathrm{x})} \mathrm{dx}=\log (\mathrm{f}(\mathrm{x}))+\mathrm{C}\) and Put, \(\quad(\mathrm{x}+\mathrm{y})=\mathrm{t} \Rightarrow\left(1+\frac{\mathrm{dy}}{\mathrm{dx}}\right)=\mathrm{dt}\) \(\log (\mathrm{xy})=\int \mathrm{t}^{-2} \mathrm{dt} \Rightarrow \log (\mathrm{xy})=\frac{\mathrm{t}^{-1}}{-1}+\mathrm{c}\) \(\log (x y)=\frac{(x+y)^{-1}}{-1}+C\) \(\log (x y)=\frac{-1}{x+y}+C\)
COMEDK-2014
Differential Equation
87206
The differential equation of \(y=\mathbf{a e}^{\mathbf{b x + c}}\) is
(C) : Given that the differential equation \(y=a e^{b x+c} \tag{i}\) Differentiating equation (i) w.r.t. \(x\), we get - \(y_{1}=a\left(e^{b x+c}\right)(b)=b\left(a e^{b x+c}\right)=b y\) \(b=\frac{y_{1}}{y} \tag{ii}\) Again differentiating equation (ii) w.r.t. \(\mathrm{x}\), we get - \(0=\frac{y \cdot y_{2}-y_{1} \cdot y_{1}}{y^{2}} \Rightarrow y \cdot y_{2}-y_{1} y_{1}=0\) \(y_{1}^{2}=y \cdot y_{2}\)