87195
The equation of the curve passing through the point \(\left(a,-\frac{1}{a}\right)\) and satisfying the differential equation \(y-x \frac{d y}{d x}=a\left(y^{2}+\frac{d y}{d x}\right)\) is
1 \((x+a)(1+a y)=-4 a^{2} y\)
2 \((x+a)(1-a y)=4 a^{2} y\)
3 \((x+a)(1-a y)=-4 a^{2} y\)
4 None of these
Explanation:
(C) : We have, \(y-x \frac{d y}{d x}=a\left(y^{2}+\frac{d y}{d x}\right)\) \(y d x-x d y=a y^{2} d x+a d y\) \(y(1-a y) d x=(x+a) d y\) \(\frac{d x}{x+a}-\frac{d y}{y(1-a y)}=0\) On integrating we get, \(\log (\mathrm{x}+\mathrm{a})-\log \mathrm{y}+\log (1-\mathrm{ay})=\log \mathrm{C}\) or \(\log \frac{(a+x)(1-a y)}{y}=\log C\) \((x+a)(1-a y)=C y\) Since, the curve passes through \(\left(\mathrm{a},-\frac{1}{\mathrm{a}}\right)\) \(\therefore \quad 2 \mathrm{a} \times(1+1)=-\frac{\mathrm{C}}{\mathrm{a}} \Rightarrow \mathrm{C}=-4 \mathrm{a}^{2}\) So, \(\quad(x+a)(1-a y)=-4 a^{2} y\)
BITSAT-2019
Differential Equation
87196
The general solution of differential equation \(\left(e^{x}+1\right) y d y=(y+1) e^{x} d x\) is
(C) : Given that, \(\left(e^{x}+1\right) y d y=(y+1) e^{x} d x\) \(\frac{d x}{d y}=\frac{y}{1+y}+\frac{y}{(1+y) e^{x}} \Rightarrow \frac{d x}{d y}=\left(\frac{y}{1+y}\right)\left(1+\frac{1}{e^{x}}\right)\) \(\frac{d x}{d y}=\left(\frac{y}{1+y}\right)\left(\frac{e^{x}+1}{e^{x}}\right) \Rightarrow\left(\frac{y}{1+y}\right) d y=\left(\frac{e^{x}}{e^{x}+1}\right) d x\) After integrating on both sides, we get - \(\int \frac{\mathrm{y}}{1+\mathrm{y}} \mathrm{dy}=\int \frac{\mathrm{e}^{\mathrm{x}}}{1+\mathrm{e}^{\mathrm{x}}} \mathrm{dx}\) \(\mathrm{y}-\log |(1+\mathrm{y})|=\log \left|\left(1+\mathrm{e}^{\mathrm{x}}\right)\right|+\log \mathrm{k}\) Hence, \(\mathrm{y}=\log \left[\mathrm{k}(1+\mathrm{y})\left(1+\mathrm{e}^{\mathrm{x}}\right)\right]\)
BITSAT-2019
Differential Equation
87197
The solutions of \((x+y+1) d y=d x\) are
1 \(x+y+2=\mathrm{Ce}^{y}\)
2 \(x+y+4=C \log y\)
3 \(\log (x+y+2)=C y\)
4 \(\log (x+y+2)=C-y\)
Explanation:
(A) : We have differential equation - \((x+y+1) d y=d x\) \(\frac{d x}{d y}=x+y+1 \Rightarrow \frac{d x}{d y}-x=y+1 \tag{i}\) Comparing the equation we get - \(\mathrm{P}=-1\) and \(\mathrm{Q}=\mathrm{y}+1\) \(I . F=\mathrm{e}^{\int P \mathrm{dy}} \Rightarrow \quad \mathrm{I} . F=\mathrm{e}^{\int-1 \mathrm{dy}}=\mathrm{e}^{-\mathrm{y}}\) Now general solution we get - x. I. \(F=\int\) Q.(I.F)dy + C x. \(e^{-y}=\int e^{-y} \cdot(y+1) d y+C\) \(x \cdot e^{-y}=(y+1) \int e^{-y} d y-\int\left(\frac{d}{d y}(y+1) \cdot \int e^{-y} d y\right) d y\) \(=-(y+1) \cdot e^{-y}-\int(1)\left(-e^{-y}\right) d y+C\) \(\mathrm{x} \mathrm{e}^{-\mathrm{y}}=-(\mathrm{y}+1) \mathrm{e}^{-\mathrm{y}}-\mathrm{e}^{-\mathrm{y}}+\mathrm{C}\) \(x \mathrm{e}^{-\mathrm{y}}=\mathrm{C}-\mathrm{e}^{-\mathrm{y}}(\mathrm{y}+2)\) or \(\quad \mathrm{x}+\mathrm{y}+2=\mathrm{Ce}^{\mathrm{y}}\)
BITSAT-2020
Differential Equation
87198
The solution of differential equation \(\frac{d y}{d x}-y \tan x=-y^{2} \sec x\) is
87195
The equation of the curve passing through the point \(\left(a,-\frac{1}{a}\right)\) and satisfying the differential equation \(y-x \frac{d y}{d x}=a\left(y^{2}+\frac{d y}{d x}\right)\) is
1 \((x+a)(1+a y)=-4 a^{2} y\)
2 \((x+a)(1-a y)=4 a^{2} y\)
3 \((x+a)(1-a y)=-4 a^{2} y\)
4 None of these
Explanation:
(C) : We have, \(y-x \frac{d y}{d x}=a\left(y^{2}+\frac{d y}{d x}\right)\) \(y d x-x d y=a y^{2} d x+a d y\) \(y(1-a y) d x=(x+a) d y\) \(\frac{d x}{x+a}-\frac{d y}{y(1-a y)}=0\) On integrating we get, \(\log (\mathrm{x}+\mathrm{a})-\log \mathrm{y}+\log (1-\mathrm{ay})=\log \mathrm{C}\) or \(\log \frac{(a+x)(1-a y)}{y}=\log C\) \((x+a)(1-a y)=C y\) Since, the curve passes through \(\left(\mathrm{a},-\frac{1}{\mathrm{a}}\right)\) \(\therefore \quad 2 \mathrm{a} \times(1+1)=-\frac{\mathrm{C}}{\mathrm{a}} \Rightarrow \mathrm{C}=-4 \mathrm{a}^{2}\) So, \(\quad(x+a)(1-a y)=-4 a^{2} y\)
BITSAT-2019
Differential Equation
87196
The general solution of differential equation \(\left(e^{x}+1\right) y d y=(y+1) e^{x} d x\) is
(C) : Given that, \(\left(e^{x}+1\right) y d y=(y+1) e^{x} d x\) \(\frac{d x}{d y}=\frac{y}{1+y}+\frac{y}{(1+y) e^{x}} \Rightarrow \frac{d x}{d y}=\left(\frac{y}{1+y}\right)\left(1+\frac{1}{e^{x}}\right)\) \(\frac{d x}{d y}=\left(\frac{y}{1+y}\right)\left(\frac{e^{x}+1}{e^{x}}\right) \Rightarrow\left(\frac{y}{1+y}\right) d y=\left(\frac{e^{x}}{e^{x}+1}\right) d x\) After integrating on both sides, we get - \(\int \frac{\mathrm{y}}{1+\mathrm{y}} \mathrm{dy}=\int \frac{\mathrm{e}^{\mathrm{x}}}{1+\mathrm{e}^{\mathrm{x}}} \mathrm{dx}\) \(\mathrm{y}-\log |(1+\mathrm{y})|=\log \left|\left(1+\mathrm{e}^{\mathrm{x}}\right)\right|+\log \mathrm{k}\) Hence, \(\mathrm{y}=\log \left[\mathrm{k}(1+\mathrm{y})\left(1+\mathrm{e}^{\mathrm{x}}\right)\right]\)
BITSAT-2019
Differential Equation
87197
The solutions of \((x+y+1) d y=d x\) are
1 \(x+y+2=\mathrm{Ce}^{y}\)
2 \(x+y+4=C \log y\)
3 \(\log (x+y+2)=C y\)
4 \(\log (x+y+2)=C-y\)
Explanation:
(A) : We have differential equation - \((x+y+1) d y=d x\) \(\frac{d x}{d y}=x+y+1 \Rightarrow \frac{d x}{d y}-x=y+1 \tag{i}\) Comparing the equation we get - \(\mathrm{P}=-1\) and \(\mathrm{Q}=\mathrm{y}+1\) \(I . F=\mathrm{e}^{\int P \mathrm{dy}} \Rightarrow \quad \mathrm{I} . F=\mathrm{e}^{\int-1 \mathrm{dy}}=\mathrm{e}^{-\mathrm{y}}\) Now general solution we get - x. I. \(F=\int\) Q.(I.F)dy + C x. \(e^{-y}=\int e^{-y} \cdot(y+1) d y+C\) \(x \cdot e^{-y}=(y+1) \int e^{-y} d y-\int\left(\frac{d}{d y}(y+1) \cdot \int e^{-y} d y\right) d y\) \(=-(y+1) \cdot e^{-y}-\int(1)\left(-e^{-y}\right) d y+C\) \(\mathrm{x} \mathrm{e}^{-\mathrm{y}}=-(\mathrm{y}+1) \mathrm{e}^{-\mathrm{y}}-\mathrm{e}^{-\mathrm{y}}+\mathrm{C}\) \(x \mathrm{e}^{-\mathrm{y}}=\mathrm{C}-\mathrm{e}^{-\mathrm{y}}(\mathrm{y}+2)\) or \(\quad \mathrm{x}+\mathrm{y}+2=\mathrm{Ce}^{\mathrm{y}}\)
BITSAT-2020
Differential Equation
87198
The solution of differential equation \(\frac{d y}{d x}-y \tan x=-y^{2} \sec x\) is
87195
The equation of the curve passing through the point \(\left(a,-\frac{1}{a}\right)\) and satisfying the differential equation \(y-x \frac{d y}{d x}=a\left(y^{2}+\frac{d y}{d x}\right)\) is
1 \((x+a)(1+a y)=-4 a^{2} y\)
2 \((x+a)(1-a y)=4 a^{2} y\)
3 \((x+a)(1-a y)=-4 a^{2} y\)
4 None of these
Explanation:
(C) : We have, \(y-x \frac{d y}{d x}=a\left(y^{2}+\frac{d y}{d x}\right)\) \(y d x-x d y=a y^{2} d x+a d y\) \(y(1-a y) d x=(x+a) d y\) \(\frac{d x}{x+a}-\frac{d y}{y(1-a y)}=0\) On integrating we get, \(\log (\mathrm{x}+\mathrm{a})-\log \mathrm{y}+\log (1-\mathrm{ay})=\log \mathrm{C}\) or \(\log \frac{(a+x)(1-a y)}{y}=\log C\) \((x+a)(1-a y)=C y\) Since, the curve passes through \(\left(\mathrm{a},-\frac{1}{\mathrm{a}}\right)\) \(\therefore \quad 2 \mathrm{a} \times(1+1)=-\frac{\mathrm{C}}{\mathrm{a}} \Rightarrow \mathrm{C}=-4 \mathrm{a}^{2}\) So, \(\quad(x+a)(1-a y)=-4 a^{2} y\)
BITSAT-2019
Differential Equation
87196
The general solution of differential equation \(\left(e^{x}+1\right) y d y=(y+1) e^{x} d x\) is
(C) : Given that, \(\left(e^{x}+1\right) y d y=(y+1) e^{x} d x\) \(\frac{d x}{d y}=\frac{y}{1+y}+\frac{y}{(1+y) e^{x}} \Rightarrow \frac{d x}{d y}=\left(\frac{y}{1+y}\right)\left(1+\frac{1}{e^{x}}\right)\) \(\frac{d x}{d y}=\left(\frac{y}{1+y}\right)\left(\frac{e^{x}+1}{e^{x}}\right) \Rightarrow\left(\frac{y}{1+y}\right) d y=\left(\frac{e^{x}}{e^{x}+1}\right) d x\) After integrating on both sides, we get - \(\int \frac{\mathrm{y}}{1+\mathrm{y}} \mathrm{dy}=\int \frac{\mathrm{e}^{\mathrm{x}}}{1+\mathrm{e}^{\mathrm{x}}} \mathrm{dx}\) \(\mathrm{y}-\log |(1+\mathrm{y})|=\log \left|\left(1+\mathrm{e}^{\mathrm{x}}\right)\right|+\log \mathrm{k}\) Hence, \(\mathrm{y}=\log \left[\mathrm{k}(1+\mathrm{y})\left(1+\mathrm{e}^{\mathrm{x}}\right)\right]\)
BITSAT-2019
Differential Equation
87197
The solutions of \((x+y+1) d y=d x\) are
1 \(x+y+2=\mathrm{Ce}^{y}\)
2 \(x+y+4=C \log y\)
3 \(\log (x+y+2)=C y\)
4 \(\log (x+y+2)=C-y\)
Explanation:
(A) : We have differential equation - \((x+y+1) d y=d x\) \(\frac{d x}{d y}=x+y+1 \Rightarrow \frac{d x}{d y}-x=y+1 \tag{i}\) Comparing the equation we get - \(\mathrm{P}=-1\) and \(\mathrm{Q}=\mathrm{y}+1\) \(I . F=\mathrm{e}^{\int P \mathrm{dy}} \Rightarrow \quad \mathrm{I} . F=\mathrm{e}^{\int-1 \mathrm{dy}}=\mathrm{e}^{-\mathrm{y}}\) Now general solution we get - x. I. \(F=\int\) Q.(I.F)dy + C x. \(e^{-y}=\int e^{-y} \cdot(y+1) d y+C\) \(x \cdot e^{-y}=(y+1) \int e^{-y} d y-\int\left(\frac{d}{d y}(y+1) \cdot \int e^{-y} d y\right) d y\) \(=-(y+1) \cdot e^{-y}-\int(1)\left(-e^{-y}\right) d y+C\) \(\mathrm{x} \mathrm{e}^{-\mathrm{y}}=-(\mathrm{y}+1) \mathrm{e}^{-\mathrm{y}}-\mathrm{e}^{-\mathrm{y}}+\mathrm{C}\) \(x \mathrm{e}^{-\mathrm{y}}=\mathrm{C}-\mathrm{e}^{-\mathrm{y}}(\mathrm{y}+2)\) or \(\quad \mathrm{x}+\mathrm{y}+2=\mathrm{Ce}^{\mathrm{y}}\)
BITSAT-2020
Differential Equation
87198
The solution of differential equation \(\frac{d y}{d x}-y \tan x=-y^{2} \sec x\) is
87195
The equation of the curve passing through the point \(\left(a,-\frac{1}{a}\right)\) and satisfying the differential equation \(y-x \frac{d y}{d x}=a\left(y^{2}+\frac{d y}{d x}\right)\) is
1 \((x+a)(1+a y)=-4 a^{2} y\)
2 \((x+a)(1-a y)=4 a^{2} y\)
3 \((x+a)(1-a y)=-4 a^{2} y\)
4 None of these
Explanation:
(C) : We have, \(y-x \frac{d y}{d x}=a\left(y^{2}+\frac{d y}{d x}\right)\) \(y d x-x d y=a y^{2} d x+a d y\) \(y(1-a y) d x=(x+a) d y\) \(\frac{d x}{x+a}-\frac{d y}{y(1-a y)}=0\) On integrating we get, \(\log (\mathrm{x}+\mathrm{a})-\log \mathrm{y}+\log (1-\mathrm{ay})=\log \mathrm{C}\) or \(\log \frac{(a+x)(1-a y)}{y}=\log C\) \((x+a)(1-a y)=C y\) Since, the curve passes through \(\left(\mathrm{a},-\frac{1}{\mathrm{a}}\right)\) \(\therefore \quad 2 \mathrm{a} \times(1+1)=-\frac{\mathrm{C}}{\mathrm{a}} \Rightarrow \mathrm{C}=-4 \mathrm{a}^{2}\) So, \(\quad(x+a)(1-a y)=-4 a^{2} y\)
BITSAT-2019
Differential Equation
87196
The general solution of differential equation \(\left(e^{x}+1\right) y d y=(y+1) e^{x} d x\) is
(C) : Given that, \(\left(e^{x}+1\right) y d y=(y+1) e^{x} d x\) \(\frac{d x}{d y}=\frac{y}{1+y}+\frac{y}{(1+y) e^{x}} \Rightarrow \frac{d x}{d y}=\left(\frac{y}{1+y}\right)\left(1+\frac{1}{e^{x}}\right)\) \(\frac{d x}{d y}=\left(\frac{y}{1+y}\right)\left(\frac{e^{x}+1}{e^{x}}\right) \Rightarrow\left(\frac{y}{1+y}\right) d y=\left(\frac{e^{x}}{e^{x}+1}\right) d x\) After integrating on both sides, we get - \(\int \frac{\mathrm{y}}{1+\mathrm{y}} \mathrm{dy}=\int \frac{\mathrm{e}^{\mathrm{x}}}{1+\mathrm{e}^{\mathrm{x}}} \mathrm{dx}\) \(\mathrm{y}-\log |(1+\mathrm{y})|=\log \left|\left(1+\mathrm{e}^{\mathrm{x}}\right)\right|+\log \mathrm{k}\) Hence, \(\mathrm{y}=\log \left[\mathrm{k}(1+\mathrm{y})\left(1+\mathrm{e}^{\mathrm{x}}\right)\right]\)
BITSAT-2019
Differential Equation
87197
The solutions of \((x+y+1) d y=d x\) are
1 \(x+y+2=\mathrm{Ce}^{y}\)
2 \(x+y+4=C \log y\)
3 \(\log (x+y+2)=C y\)
4 \(\log (x+y+2)=C-y\)
Explanation:
(A) : We have differential equation - \((x+y+1) d y=d x\) \(\frac{d x}{d y}=x+y+1 \Rightarrow \frac{d x}{d y}-x=y+1 \tag{i}\) Comparing the equation we get - \(\mathrm{P}=-1\) and \(\mathrm{Q}=\mathrm{y}+1\) \(I . F=\mathrm{e}^{\int P \mathrm{dy}} \Rightarrow \quad \mathrm{I} . F=\mathrm{e}^{\int-1 \mathrm{dy}}=\mathrm{e}^{-\mathrm{y}}\) Now general solution we get - x. I. \(F=\int\) Q.(I.F)dy + C x. \(e^{-y}=\int e^{-y} \cdot(y+1) d y+C\) \(x \cdot e^{-y}=(y+1) \int e^{-y} d y-\int\left(\frac{d}{d y}(y+1) \cdot \int e^{-y} d y\right) d y\) \(=-(y+1) \cdot e^{-y}-\int(1)\left(-e^{-y}\right) d y+C\) \(\mathrm{x} \mathrm{e}^{-\mathrm{y}}=-(\mathrm{y}+1) \mathrm{e}^{-\mathrm{y}}-\mathrm{e}^{-\mathrm{y}}+\mathrm{C}\) \(x \mathrm{e}^{-\mathrm{y}}=\mathrm{C}-\mathrm{e}^{-\mathrm{y}}(\mathrm{y}+2)\) or \(\quad \mathrm{x}+\mathrm{y}+2=\mathrm{Ce}^{\mathrm{y}}\)
BITSAT-2020
Differential Equation
87198
The solution of differential equation \(\frac{d y}{d x}-y \tan x=-y^{2} \sec x\) is
