(B) : Given, \(\frac{\mathrm{dy}}{\mathrm{dx}}+\frac{1}{3} \mathrm{y}=1\) Hence, \(\mathrm{P}=\frac{1}{3}\) and \(\mathrm{Q}=1\) \(\mathrm{I} . F=\mathrm{e}^{\int \mathrm{pdx}} \Rightarrow \mathrm{I} . F=\mathrm{e}^{\int \frac{1}{3} \mathrm{dx}}\) \(\text { I.F }=\mathrm{e}^{\frac{1}{3} \mathrm{x}}\) Solution of general equation, \(y \cdot e^{x / 3}=\int e^{x / 3} d x \Rightarrow y \cdot e^{x / 3}=\frac{e^{x / 3}}{1 / 3}+c\) \(y \cdot e^{x / 3}=3 e^{x / 3}+c\) \(y=3+c e^{-x / 3}\)
AP EAMCET-2002
Differential Equation
87172
The solution for the differential equation
\(\frac{dy}{y} + \frac{dx}{x} = 0\) is
1 \(\frac{1}{y} + \frac{1}{x} = c\)
2 \(\log x \cdot \log y = c\)
3 \(xy = c\)
4 \(x + y = c\) Ans. (c) Exp: (C) : We have differential equation, \(\sqrt{1-x^{2} y^{2}} d x=y d x+x d y \Rightarrow \sqrt{1-(x y)^{2}} d x=d(x y)\) \(\mathrm{dx}=\frac{\mathrm{d}(\mathrm{xy})}{\sqrt{1-(\mathrm{xy})^{2}}}\) Let, \(\quad \begin{array}{ll}x y=t \\ d(x y)=d t\end{array}\) Integrating on both sides, \(\int d x=\frac{d t}{\sqrt{1-t^{2}}} \Rightarrow x=\sin ^{-1} t+c\) \(x=\sin ^{-1}(x y)+c \Rightarrow(x+c)=\sin ^{-1}(x y)\) \(\sin (x+c)=x y\)
Explanation:
(C) : We have differential equation, \(\sqrt{1-x^{2} y^{2}} d x=y d x+x d y \Rightarrow \sqrt{1-(x y)^{2}} d x=d(x y)\) \(\mathrm{dx}=\frac{\mathrm{d}(\mathrm{xy})}{\sqrt{1-(\mathrm{xy})^{2}}}\) Let, \(\quad \begin{array}{ll}x y=t \\ d(x y)=d t\end{array}\) Integrating on both sides, \(\int d x=\frac{d t}{\sqrt{1-t^{2}}} \Rightarrow x=\sin ^{-1} t+c\) \(x=\sin ^{-1}(x y)+c \Rightarrow(x+c)=\sin ^{-1}(x y)\) \(\sin (x+c)=x y\)
Karnataka CET-2016
Differential Equation
87182
The general solution of the differential equation of all circles having centre at \(A(-1,2)\) is
(D) : We have differential equation of all circle \(x^{2}+y^{2}+2 g x+2 f y+c=0\) Having centre \((-1,2) \Rightarrow(-\mathrm{g},-\mathrm{f})\) \(x^{2}+y^{2}+2 g x+2 f y+c=0\) Having centre \((-1,2) \Rightarrow(-\mathrm{g},-\mathrm{f})\) \(\mathrm{x}^{2}+\mathrm{y}^{2}+2 \times 1 \times \mathrm{x}+2 \times(-2) \times \mathrm{y}+\mathrm{c}=0\) Equate of circle, \(x^{2}+y^{2}+2 x-4 y+c=0\)
MHT CET-2019
Differential Equation
87183
The particular solution of the differential equation \(x d y+2 y d x=0\), when \(x=2, y=1\) is
1 \(x y=4\)
2 \(x^{2} y=4\)
3 \(x y^{2}=4\)
4 \(x^{2} y^{2}=4\)
Explanation:
(B) : Given the differential equation, \(x d y+2 y d x=0 \Rightarrow x d y=-2 y d x\) \(\frac{d y}{y}=\frac{-2 d x}{x}\) Integrating both sides we get, \(\log y=-2 \log x+\log c \Rightarrow \log y+2 \log x=\log c\) \(x^{2} y=c\) When, \(x=2\) and \(y=1\) Then, \((2)^{2} \times 1=\mathrm{c} \Rightarrow \mathrm{c}=4\) Then, \(x^{2} y=4\)
(B) : Given, \(\frac{\mathrm{dy}}{\mathrm{dx}}+\frac{1}{3} \mathrm{y}=1\) Hence, \(\mathrm{P}=\frac{1}{3}\) and \(\mathrm{Q}=1\) \(\mathrm{I} . F=\mathrm{e}^{\int \mathrm{pdx}} \Rightarrow \mathrm{I} . F=\mathrm{e}^{\int \frac{1}{3} \mathrm{dx}}\) \(\text { I.F }=\mathrm{e}^{\frac{1}{3} \mathrm{x}}\) Solution of general equation, \(y \cdot e^{x / 3}=\int e^{x / 3} d x \Rightarrow y \cdot e^{x / 3}=\frac{e^{x / 3}}{1 / 3}+c\) \(y \cdot e^{x / 3}=3 e^{x / 3}+c\) \(y=3+c e^{-x / 3}\)
AP EAMCET-2002
Differential Equation
87172
The solution for the differential equation
\(\frac{dy}{y} + \frac{dx}{x} = 0\) is
1 \(\frac{1}{y} + \frac{1}{x} = c\)
2 \(\log x \cdot \log y = c\)
3 \(xy = c\)
4 \(x + y = c\) Ans. (c) Exp: (C) : We have differential equation, \(\sqrt{1-x^{2} y^{2}} d x=y d x+x d y \Rightarrow \sqrt{1-(x y)^{2}} d x=d(x y)\) \(\mathrm{dx}=\frac{\mathrm{d}(\mathrm{xy})}{\sqrt{1-(\mathrm{xy})^{2}}}\) Let, \(\quad \begin{array}{ll}x y=t \\ d(x y)=d t\end{array}\) Integrating on both sides, \(\int d x=\frac{d t}{\sqrt{1-t^{2}}} \Rightarrow x=\sin ^{-1} t+c\) \(x=\sin ^{-1}(x y)+c \Rightarrow(x+c)=\sin ^{-1}(x y)\) \(\sin (x+c)=x y\)
Explanation:
(C) : We have differential equation, \(\sqrt{1-x^{2} y^{2}} d x=y d x+x d y \Rightarrow \sqrt{1-(x y)^{2}} d x=d(x y)\) \(\mathrm{dx}=\frac{\mathrm{d}(\mathrm{xy})}{\sqrt{1-(\mathrm{xy})^{2}}}\) Let, \(\quad \begin{array}{ll}x y=t \\ d(x y)=d t\end{array}\) Integrating on both sides, \(\int d x=\frac{d t}{\sqrt{1-t^{2}}} \Rightarrow x=\sin ^{-1} t+c\) \(x=\sin ^{-1}(x y)+c \Rightarrow(x+c)=\sin ^{-1}(x y)\) \(\sin (x+c)=x y\)
Karnataka CET-2016
Differential Equation
87182
The general solution of the differential equation of all circles having centre at \(A(-1,2)\) is
(D) : We have differential equation of all circle \(x^{2}+y^{2}+2 g x+2 f y+c=0\) Having centre \((-1,2) \Rightarrow(-\mathrm{g},-\mathrm{f})\) \(x^{2}+y^{2}+2 g x+2 f y+c=0\) Having centre \((-1,2) \Rightarrow(-\mathrm{g},-\mathrm{f})\) \(\mathrm{x}^{2}+\mathrm{y}^{2}+2 \times 1 \times \mathrm{x}+2 \times(-2) \times \mathrm{y}+\mathrm{c}=0\) Equate of circle, \(x^{2}+y^{2}+2 x-4 y+c=0\)
MHT CET-2019
Differential Equation
87183
The particular solution of the differential equation \(x d y+2 y d x=0\), when \(x=2, y=1\) is
1 \(x y=4\)
2 \(x^{2} y=4\)
3 \(x y^{2}=4\)
4 \(x^{2} y^{2}=4\)
Explanation:
(B) : Given the differential equation, \(x d y+2 y d x=0 \Rightarrow x d y=-2 y d x\) \(\frac{d y}{y}=\frac{-2 d x}{x}\) Integrating both sides we get, \(\log y=-2 \log x+\log c \Rightarrow \log y+2 \log x=\log c\) \(x^{2} y=c\) When, \(x=2\) and \(y=1\) Then, \((2)^{2} \times 1=\mathrm{c} \Rightarrow \mathrm{c}=4\) Then, \(x^{2} y=4\)
(B) : Given, \(\frac{\mathrm{dy}}{\mathrm{dx}}+\frac{1}{3} \mathrm{y}=1\) Hence, \(\mathrm{P}=\frac{1}{3}\) and \(\mathrm{Q}=1\) \(\mathrm{I} . F=\mathrm{e}^{\int \mathrm{pdx}} \Rightarrow \mathrm{I} . F=\mathrm{e}^{\int \frac{1}{3} \mathrm{dx}}\) \(\text { I.F }=\mathrm{e}^{\frac{1}{3} \mathrm{x}}\) Solution of general equation, \(y \cdot e^{x / 3}=\int e^{x / 3} d x \Rightarrow y \cdot e^{x / 3}=\frac{e^{x / 3}}{1 / 3}+c\) \(y \cdot e^{x / 3}=3 e^{x / 3}+c\) \(y=3+c e^{-x / 3}\)
AP EAMCET-2002
Differential Equation
87172
The solution for the differential equation
\(\frac{dy}{y} + \frac{dx}{x} = 0\) is
1 \(\frac{1}{y} + \frac{1}{x} = c\)
2 \(\log x \cdot \log y = c\)
3 \(xy = c\)
4 \(x + y = c\) Ans. (c) Exp: (C) : We have differential equation, \(\sqrt{1-x^{2} y^{2}} d x=y d x+x d y \Rightarrow \sqrt{1-(x y)^{2}} d x=d(x y)\) \(\mathrm{dx}=\frac{\mathrm{d}(\mathrm{xy})}{\sqrt{1-(\mathrm{xy})^{2}}}\) Let, \(\quad \begin{array}{ll}x y=t \\ d(x y)=d t\end{array}\) Integrating on both sides, \(\int d x=\frac{d t}{\sqrt{1-t^{2}}} \Rightarrow x=\sin ^{-1} t+c\) \(x=\sin ^{-1}(x y)+c \Rightarrow(x+c)=\sin ^{-1}(x y)\) \(\sin (x+c)=x y\)
Explanation:
(C) : We have differential equation, \(\sqrt{1-x^{2} y^{2}} d x=y d x+x d y \Rightarrow \sqrt{1-(x y)^{2}} d x=d(x y)\) \(\mathrm{dx}=\frac{\mathrm{d}(\mathrm{xy})}{\sqrt{1-(\mathrm{xy})^{2}}}\) Let, \(\quad \begin{array}{ll}x y=t \\ d(x y)=d t\end{array}\) Integrating on both sides, \(\int d x=\frac{d t}{\sqrt{1-t^{2}}} \Rightarrow x=\sin ^{-1} t+c\) \(x=\sin ^{-1}(x y)+c \Rightarrow(x+c)=\sin ^{-1}(x y)\) \(\sin (x+c)=x y\)
Karnataka CET-2016
Differential Equation
87182
The general solution of the differential equation of all circles having centre at \(A(-1,2)\) is
(D) : We have differential equation of all circle \(x^{2}+y^{2}+2 g x+2 f y+c=0\) Having centre \((-1,2) \Rightarrow(-\mathrm{g},-\mathrm{f})\) \(x^{2}+y^{2}+2 g x+2 f y+c=0\) Having centre \((-1,2) \Rightarrow(-\mathrm{g},-\mathrm{f})\) \(\mathrm{x}^{2}+\mathrm{y}^{2}+2 \times 1 \times \mathrm{x}+2 \times(-2) \times \mathrm{y}+\mathrm{c}=0\) Equate of circle, \(x^{2}+y^{2}+2 x-4 y+c=0\)
MHT CET-2019
Differential Equation
87183
The particular solution of the differential equation \(x d y+2 y d x=0\), when \(x=2, y=1\) is
1 \(x y=4\)
2 \(x^{2} y=4\)
3 \(x y^{2}=4\)
4 \(x^{2} y^{2}=4\)
Explanation:
(B) : Given the differential equation, \(x d y+2 y d x=0 \Rightarrow x d y=-2 y d x\) \(\frac{d y}{y}=\frac{-2 d x}{x}\) Integrating both sides we get, \(\log y=-2 \log x+\log c \Rightarrow \log y+2 \log x=\log c\) \(x^{2} y=c\) When, \(x=2\) and \(y=1\) Then, \((2)^{2} \times 1=\mathrm{c} \Rightarrow \mathrm{c}=4\) Then, \(x^{2} y=4\)
(B) : Given, \(\frac{\mathrm{dy}}{\mathrm{dx}}+\frac{1}{3} \mathrm{y}=1\) Hence, \(\mathrm{P}=\frac{1}{3}\) and \(\mathrm{Q}=1\) \(\mathrm{I} . F=\mathrm{e}^{\int \mathrm{pdx}} \Rightarrow \mathrm{I} . F=\mathrm{e}^{\int \frac{1}{3} \mathrm{dx}}\) \(\text { I.F }=\mathrm{e}^{\frac{1}{3} \mathrm{x}}\) Solution of general equation, \(y \cdot e^{x / 3}=\int e^{x / 3} d x \Rightarrow y \cdot e^{x / 3}=\frac{e^{x / 3}}{1 / 3}+c\) \(y \cdot e^{x / 3}=3 e^{x / 3}+c\) \(y=3+c e^{-x / 3}\)
AP EAMCET-2002
Differential Equation
87172
The solution for the differential equation
\(\frac{dy}{y} + \frac{dx}{x} = 0\) is
1 \(\frac{1}{y} + \frac{1}{x} = c\)
2 \(\log x \cdot \log y = c\)
3 \(xy = c\)
4 \(x + y = c\) Ans. (c) Exp: (C) : We have differential equation, \(\sqrt{1-x^{2} y^{2}} d x=y d x+x d y \Rightarrow \sqrt{1-(x y)^{2}} d x=d(x y)\) \(\mathrm{dx}=\frac{\mathrm{d}(\mathrm{xy})}{\sqrt{1-(\mathrm{xy})^{2}}}\) Let, \(\quad \begin{array}{ll}x y=t \\ d(x y)=d t\end{array}\) Integrating on both sides, \(\int d x=\frac{d t}{\sqrt{1-t^{2}}} \Rightarrow x=\sin ^{-1} t+c\) \(x=\sin ^{-1}(x y)+c \Rightarrow(x+c)=\sin ^{-1}(x y)\) \(\sin (x+c)=x y\)
Explanation:
(C) : We have differential equation, \(\sqrt{1-x^{2} y^{2}} d x=y d x+x d y \Rightarrow \sqrt{1-(x y)^{2}} d x=d(x y)\) \(\mathrm{dx}=\frac{\mathrm{d}(\mathrm{xy})}{\sqrt{1-(\mathrm{xy})^{2}}}\) Let, \(\quad \begin{array}{ll}x y=t \\ d(x y)=d t\end{array}\) Integrating on both sides, \(\int d x=\frac{d t}{\sqrt{1-t^{2}}} \Rightarrow x=\sin ^{-1} t+c\) \(x=\sin ^{-1}(x y)+c \Rightarrow(x+c)=\sin ^{-1}(x y)\) \(\sin (x+c)=x y\)
Karnataka CET-2016
Differential Equation
87182
The general solution of the differential equation of all circles having centre at \(A(-1,2)\) is
(D) : We have differential equation of all circle \(x^{2}+y^{2}+2 g x+2 f y+c=0\) Having centre \((-1,2) \Rightarrow(-\mathrm{g},-\mathrm{f})\) \(x^{2}+y^{2}+2 g x+2 f y+c=0\) Having centre \((-1,2) \Rightarrow(-\mathrm{g},-\mathrm{f})\) \(\mathrm{x}^{2}+\mathrm{y}^{2}+2 \times 1 \times \mathrm{x}+2 \times(-2) \times \mathrm{y}+\mathrm{c}=0\) Equate of circle, \(x^{2}+y^{2}+2 x-4 y+c=0\)
MHT CET-2019
Differential Equation
87183
The particular solution of the differential equation \(x d y+2 y d x=0\), when \(x=2, y=1\) is
1 \(x y=4\)
2 \(x^{2} y=4\)
3 \(x y^{2}=4\)
4 \(x^{2} y^{2}=4\)
Explanation:
(B) : Given the differential equation, \(x d y+2 y d x=0 \Rightarrow x d y=-2 y d x\) \(\frac{d y}{y}=\frac{-2 d x}{x}\) Integrating both sides we get, \(\log y=-2 \log x+\log c \Rightarrow \log y+2 \log x=\log c\) \(x^{2} y=c\) When, \(x=2\) and \(y=1\) Then, \((2)^{2} \times 1=\mathrm{c} \Rightarrow \mathrm{c}=4\) Then, \(x^{2} y=4\)
