87554
If a curve \(y=f(x)\) passes through the point (1,1) and satisfies the differential equation, \(y(1+x y) d x=x d y\), then \(f\left(-\frac{1}{2}\right)\) is equal to
1 \(-\frac{2}{5}\)
2 \(-\frac{4}{5}\)
3 \(\frac{2}{5}\)
4 \(\frac{4}{5}\)
Explanation:
(D) : \(y(1+x y) d x=x d y\) \(y d x+x y^{2} d x=x d y\) \(\frac{x d y-y d x}{y^{2}}=x d x, \frac{-(y d x-x d y)}{y^{2}}=x d x\) \(-d\left(\frac{x}{y}\right)=x d x\) On integrating both sides, we get- \(-\frac{x}{y}=\frac{x^{2}}{2}+C \tag{i}\) \(\because\) It passes through \((1,-1)\), \(\therefore \quad 1 =\frac{1}{2}+C\) \(\mathrm{C} =\frac{1}{2}\) Now, from equation (i), \(-\frac{x}{y}=\frac{x^{2}}{2}+\frac{1}{2}\) \(x^{2}+1=-\frac{2 x}{y}\) \(y=-\frac{2 x}{x^{2}+1}\) \(\therefore \quad \mathrm{f}\left(-\frac{1}{2}\right)=\frac{4}{5}\)
JEE Main-2016]**#
Differential Equation
87535
The solution of \(\cos y+(x \sin y-1) \frac{d y}{d x}=0\) is
1 \(x\) secy \(=\tan y+C\)
2 tany - secy \(=C x\)
3 tany + secy \(=\mathrm{Cx}\)
4 \(x\) secy + tany \(=\) C
Explanation:
(A) : \(\cos \mathrm{y}+(\mathrm{x} \sin \mathrm{y}-1) \frac{\mathrm{dy}}{\mathrm{dx}}=0\) \(\frac{d x}{d y}=\frac{1-\sin y}{\cos y}\) \(\frac{d y}{d x}=\frac{\cos y}{1-x \sin y}\) \(\frac{d x}{d y}+x \tan y=\sec y\) it is a linear differential equation in y.l I.F \(=\mathrm{e}^{\int \mathrm{pdy}}=\mathrm{e}^{\int \tan y \mathrm{dy}}=\mathrm{e}^{\ln \sec \mathrm{y}}=\sec \mathrm{y}\) \(\because \quad \mathrm{x} \times \mathrm{I} . \mathrm{F}=\int \mathrm{Q} \times\) I.F. \(+\mathrm{C}\) \(\therefore \quad \mathrm{x} \sec \mathrm{y}=\int \sec \mathrm{y} \cdot \sec \mathrm{y} \mathrm{dy}=\tan \mathrm{y}+\mathrm{C}\) \(\therefore \quad x \sec y=\tan y+C\)
AP EAMCET-2014
Differential Equation
87536
Solution of the differential equation \(\cos x d y=\) \(y(\sin x-y) d x, 0\lt x\lt \frac{\pi}{2}\), is
1 \(\sec x=(\tan x+C) y\)
2 \(y \sec x=\tan x+C\)
3 \(y \tan x=\operatorname{sex} x+C\)
4 \(\tan x=(\sec x+C) y\)
Explanation:
(A) : Given, Differential equation \(\cos x d y=y(\sin x-y) d x\) \(\frac{d y}{d x}=y \tan x-y^{2} \sec x\) \(\frac{1 d y}{y^{2} d x}-\frac{1}{y} \tan x=-\sec x\) Let, \(-\frac{1}{y}=t\) \(\frac{1 d y}{y^{2} d x}=\frac{d t}{d x}\) \(\frac{d t}{d x}+t(\tan x)=-\sec x\) \(\text { I.F }=e^{\int \tan x d x}=-\sec x\) Solution is (I.F) \(=\int(\) I.F \() \sec x d x\) \(\frac{1}{y} \sec x=\tan x+c\)
AIEEE-2010
Differential Equation
87555
The curve amongst the family of curves represented by the differential equation, \(\left(x^{2}-\right.\) \(\left.\mathbf{y}^{2}\right) \mathbf{d x}+\mathbf{2 x y d y}=0\), which passes through \((1,1)\), is
1 a circle with centre on the Y-axis
2 a circle with centre on the X-axis
3 an ellipse with major axis along the \(\mathrm{Y}\)-axis
4 a hyperbola with transverse axis along the Xaxis.
Explanation:
(B) : \(\left(x^{2}-y^{2}\right) d x+2 x y d y=0\) Put, \(\quad y=v x\) \(\frac{d y}{d x}=\frac{y^{2}-x^{2}}{2 x y}\) \(y=v x\) \(\frac{d y}{d x}=v+x \frac{d v}{d x}\) Solving we get- \(\int \frac{2 v}{v^{2}+1} d v=\int-\frac{d x}{x}\) \(\ln \left(v^{2}+1\right)=-\ln x+C\) \(\mathrm{y}^{2}+\mathrm{x}^{2}=\mathrm{Cx}\) \(1+1=\mathrm{C}\) \(\mathrm{C}=2\) \(y^{2}+x^{2}=2 x\)
87554
If a curve \(y=f(x)\) passes through the point (1,1) and satisfies the differential equation, \(y(1+x y) d x=x d y\), then \(f\left(-\frac{1}{2}\right)\) is equal to
1 \(-\frac{2}{5}\)
2 \(-\frac{4}{5}\)
3 \(\frac{2}{5}\)
4 \(\frac{4}{5}\)
Explanation:
(D) : \(y(1+x y) d x=x d y\) \(y d x+x y^{2} d x=x d y\) \(\frac{x d y-y d x}{y^{2}}=x d x, \frac{-(y d x-x d y)}{y^{2}}=x d x\) \(-d\left(\frac{x}{y}\right)=x d x\) On integrating both sides, we get- \(-\frac{x}{y}=\frac{x^{2}}{2}+C \tag{i}\) \(\because\) It passes through \((1,-1)\), \(\therefore \quad 1 =\frac{1}{2}+C\) \(\mathrm{C} =\frac{1}{2}\) Now, from equation (i), \(-\frac{x}{y}=\frac{x^{2}}{2}+\frac{1}{2}\) \(x^{2}+1=-\frac{2 x}{y}\) \(y=-\frac{2 x}{x^{2}+1}\) \(\therefore \quad \mathrm{f}\left(-\frac{1}{2}\right)=\frac{4}{5}\)
JEE Main-2016]**#
Differential Equation
87535
The solution of \(\cos y+(x \sin y-1) \frac{d y}{d x}=0\) is
1 \(x\) secy \(=\tan y+C\)
2 tany - secy \(=C x\)
3 tany + secy \(=\mathrm{Cx}\)
4 \(x\) secy + tany \(=\) C
Explanation:
(A) : \(\cos \mathrm{y}+(\mathrm{x} \sin \mathrm{y}-1) \frac{\mathrm{dy}}{\mathrm{dx}}=0\) \(\frac{d x}{d y}=\frac{1-\sin y}{\cos y}\) \(\frac{d y}{d x}=\frac{\cos y}{1-x \sin y}\) \(\frac{d x}{d y}+x \tan y=\sec y\) it is a linear differential equation in y.l I.F \(=\mathrm{e}^{\int \mathrm{pdy}}=\mathrm{e}^{\int \tan y \mathrm{dy}}=\mathrm{e}^{\ln \sec \mathrm{y}}=\sec \mathrm{y}\) \(\because \quad \mathrm{x} \times \mathrm{I} . \mathrm{F}=\int \mathrm{Q} \times\) I.F. \(+\mathrm{C}\) \(\therefore \quad \mathrm{x} \sec \mathrm{y}=\int \sec \mathrm{y} \cdot \sec \mathrm{y} \mathrm{dy}=\tan \mathrm{y}+\mathrm{C}\) \(\therefore \quad x \sec y=\tan y+C\)
AP EAMCET-2014
Differential Equation
87536
Solution of the differential equation \(\cos x d y=\) \(y(\sin x-y) d x, 0\lt x\lt \frac{\pi}{2}\), is
1 \(\sec x=(\tan x+C) y\)
2 \(y \sec x=\tan x+C\)
3 \(y \tan x=\operatorname{sex} x+C\)
4 \(\tan x=(\sec x+C) y\)
Explanation:
(A) : Given, Differential equation \(\cos x d y=y(\sin x-y) d x\) \(\frac{d y}{d x}=y \tan x-y^{2} \sec x\) \(\frac{1 d y}{y^{2} d x}-\frac{1}{y} \tan x=-\sec x\) Let, \(-\frac{1}{y}=t\) \(\frac{1 d y}{y^{2} d x}=\frac{d t}{d x}\) \(\frac{d t}{d x}+t(\tan x)=-\sec x\) \(\text { I.F }=e^{\int \tan x d x}=-\sec x\) Solution is (I.F) \(=\int(\) I.F \() \sec x d x\) \(\frac{1}{y} \sec x=\tan x+c\)
AIEEE-2010
Differential Equation
87555
The curve amongst the family of curves represented by the differential equation, \(\left(x^{2}-\right.\) \(\left.\mathbf{y}^{2}\right) \mathbf{d x}+\mathbf{2 x y d y}=0\), which passes through \((1,1)\), is
1 a circle with centre on the Y-axis
2 a circle with centre on the X-axis
3 an ellipse with major axis along the \(\mathrm{Y}\)-axis
4 a hyperbola with transverse axis along the Xaxis.
Explanation:
(B) : \(\left(x^{2}-y^{2}\right) d x+2 x y d y=0\) Put, \(\quad y=v x\) \(\frac{d y}{d x}=\frac{y^{2}-x^{2}}{2 x y}\) \(y=v x\) \(\frac{d y}{d x}=v+x \frac{d v}{d x}\) Solving we get- \(\int \frac{2 v}{v^{2}+1} d v=\int-\frac{d x}{x}\) \(\ln \left(v^{2}+1\right)=-\ln x+C\) \(\mathrm{y}^{2}+\mathrm{x}^{2}=\mathrm{Cx}\) \(1+1=\mathrm{C}\) \(\mathrm{C}=2\) \(y^{2}+x^{2}=2 x\)
87554
If a curve \(y=f(x)\) passes through the point (1,1) and satisfies the differential equation, \(y(1+x y) d x=x d y\), then \(f\left(-\frac{1}{2}\right)\) is equal to
1 \(-\frac{2}{5}\)
2 \(-\frac{4}{5}\)
3 \(\frac{2}{5}\)
4 \(\frac{4}{5}\)
Explanation:
(D) : \(y(1+x y) d x=x d y\) \(y d x+x y^{2} d x=x d y\) \(\frac{x d y-y d x}{y^{2}}=x d x, \frac{-(y d x-x d y)}{y^{2}}=x d x\) \(-d\left(\frac{x}{y}\right)=x d x\) On integrating both sides, we get- \(-\frac{x}{y}=\frac{x^{2}}{2}+C \tag{i}\) \(\because\) It passes through \((1,-1)\), \(\therefore \quad 1 =\frac{1}{2}+C\) \(\mathrm{C} =\frac{1}{2}\) Now, from equation (i), \(-\frac{x}{y}=\frac{x^{2}}{2}+\frac{1}{2}\) \(x^{2}+1=-\frac{2 x}{y}\) \(y=-\frac{2 x}{x^{2}+1}\) \(\therefore \quad \mathrm{f}\left(-\frac{1}{2}\right)=\frac{4}{5}\)
JEE Main-2016]**#
Differential Equation
87535
The solution of \(\cos y+(x \sin y-1) \frac{d y}{d x}=0\) is
1 \(x\) secy \(=\tan y+C\)
2 tany - secy \(=C x\)
3 tany + secy \(=\mathrm{Cx}\)
4 \(x\) secy + tany \(=\) C
Explanation:
(A) : \(\cos \mathrm{y}+(\mathrm{x} \sin \mathrm{y}-1) \frac{\mathrm{dy}}{\mathrm{dx}}=0\) \(\frac{d x}{d y}=\frac{1-\sin y}{\cos y}\) \(\frac{d y}{d x}=\frac{\cos y}{1-x \sin y}\) \(\frac{d x}{d y}+x \tan y=\sec y\) it is a linear differential equation in y.l I.F \(=\mathrm{e}^{\int \mathrm{pdy}}=\mathrm{e}^{\int \tan y \mathrm{dy}}=\mathrm{e}^{\ln \sec \mathrm{y}}=\sec \mathrm{y}\) \(\because \quad \mathrm{x} \times \mathrm{I} . \mathrm{F}=\int \mathrm{Q} \times\) I.F. \(+\mathrm{C}\) \(\therefore \quad \mathrm{x} \sec \mathrm{y}=\int \sec \mathrm{y} \cdot \sec \mathrm{y} \mathrm{dy}=\tan \mathrm{y}+\mathrm{C}\) \(\therefore \quad x \sec y=\tan y+C\)
AP EAMCET-2014
Differential Equation
87536
Solution of the differential equation \(\cos x d y=\) \(y(\sin x-y) d x, 0\lt x\lt \frac{\pi}{2}\), is
1 \(\sec x=(\tan x+C) y\)
2 \(y \sec x=\tan x+C\)
3 \(y \tan x=\operatorname{sex} x+C\)
4 \(\tan x=(\sec x+C) y\)
Explanation:
(A) : Given, Differential equation \(\cos x d y=y(\sin x-y) d x\) \(\frac{d y}{d x}=y \tan x-y^{2} \sec x\) \(\frac{1 d y}{y^{2} d x}-\frac{1}{y} \tan x=-\sec x\) Let, \(-\frac{1}{y}=t\) \(\frac{1 d y}{y^{2} d x}=\frac{d t}{d x}\) \(\frac{d t}{d x}+t(\tan x)=-\sec x\) \(\text { I.F }=e^{\int \tan x d x}=-\sec x\) Solution is (I.F) \(=\int(\) I.F \() \sec x d x\) \(\frac{1}{y} \sec x=\tan x+c\)
AIEEE-2010
Differential Equation
87555
The curve amongst the family of curves represented by the differential equation, \(\left(x^{2}-\right.\) \(\left.\mathbf{y}^{2}\right) \mathbf{d x}+\mathbf{2 x y d y}=0\), which passes through \((1,1)\), is
1 a circle with centre on the Y-axis
2 a circle with centre on the X-axis
3 an ellipse with major axis along the \(\mathrm{Y}\)-axis
4 a hyperbola with transverse axis along the Xaxis.
Explanation:
(B) : \(\left(x^{2}-y^{2}\right) d x+2 x y d y=0\) Put, \(\quad y=v x\) \(\frac{d y}{d x}=\frac{y^{2}-x^{2}}{2 x y}\) \(y=v x\) \(\frac{d y}{d x}=v+x \frac{d v}{d x}\) Solving we get- \(\int \frac{2 v}{v^{2}+1} d v=\int-\frac{d x}{x}\) \(\ln \left(v^{2}+1\right)=-\ln x+C\) \(\mathrm{y}^{2}+\mathrm{x}^{2}=\mathrm{Cx}\) \(1+1=\mathrm{C}\) \(\mathrm{C}=2\) \(y^{2}+x^{2}=2 x\)
87554
If a curve \(y=f(x)\) passes through the point (1,1) and satisfies the differential equation, \(y(1+x y) d x=x d y\), then \(f\left(-\frac{1}{2}\right)\) is equal to
1 \(-\frac{2}{5}\)
2 \(-\frac{4}{5}\)
3 \(\frac{2}{5}\)
4 \(\frac{4}{5}\)
Explanation:
(D) : \(y(1+x y) d x=x d y\) \(y d x+x y^{2} d x=x d y\) \(\frac{x d y-y d x}{y^{2}}=x d x, \frac{-(y d x-x d y)}{y^{2}}=x d x\) \(-d\left(\frac{x}{y}\right)=x d x\) On integrating both sides, we get- \(-\frac{x}{y}=\frac{x^{2}}{2}+C \tag{i}\) \(\because\) It passes through \((1,-1)\), \(\therefore \quad 1 =\frac{1}{2}+C\) \(\mathrm{C} =\frac{1}{2}\) Now, from equation (i), \(-\frac{x}{y}=\frac{x^{2}}{2}+\frac{1}{2}\) \(x^{2}+1=-\frac{2 x}{y}\) \(y=-\frac{2 x}{x^{2}+1}\) \(\therefore \quad \mathrm{f}\left(-\frac{1}{2}\right)=\frac{4}{5}\)
JEE Main-2016]**#
Differential Equation
87535
The solution of \(\cos y+(x \sin y-1) \frac{d y}{d x}=0\) is
1 \(x\) secy \(=\tan y+C\)
2 tany - secy \(=C x\)
3 tany + secy \(=\mathrm{Cx}\)
4 \(x\) secy + tany \(=\) C
Explanation:
(A) : \(\cos \mathrm{y}+(\mathrm{x} \sin \mathrm{y}-1) \frac{\mathrm{dy}}{\mathrm{dx}}=0\) \(\frac{d x}{d y}=\frac{1-\sin y}{\cos y}\) \(\frac{d y}{d x}=\frac{\cos y}{1-x \sin y}\) \(\frac{d x}{d y}+x \tan y=\sec y\) it is a linear differential equation in y.l I.F \(=\mathrm{e}^{\int \mathrm{pdy}}=\mathrm{e}^{\int \tan y \mathrm{dy}}=\mathrm{e}^{\ln \sec \mathrm{y}}=\sec \mathrm{y}\) \(\because \quad \mathrm{x} \times \mathrm{I} . \mathrm{F}=\int \mathrm{Q} \times\) I.F. \(+\mathrm{C}\) \(\therefore \quad \mathrm{x} \sec \mathrm{y}=\int \sec \mathrm{y} \cdot \sec \mathrm{y} \mathrm{dy}=\tan \mathrm{y}+\mathrm{C}\) \(\therefore \quad x \sec y=\tan y+C\)
AP EAMCET-2014
Differential Equation
87536
Solution of the differential equation \(\cos x d y=\) \(y(\sin x-y) d x, 0\lt x\lt \frac{\pi}{2}\), is
1 \(\sec x=(\tan x+C) y\)
2 \(y \sec x=\tan x+C\)
3 \(y \tan x=\operatorname{sex} x+C\)
4 \(\tan x=(\sec x+C) y\)
Explanation:
(A) : Given, Differential equation \(\cos x d y=y(\sin x-y) d x\) \(\frac{d y}{d x}=y \tan x-y^{2} \sec x\) \(\frac{1 d y}{y^{2} d x}-\frac{1}{y} \tan x=-\sec x\) Let, \(-\frac{1}{y}=t\) \(\frac{1 d y}{y^{2} d x}=\frac{d t}{d x}\) \(\frac{d t}{d x}+t(\tan x)=-\sec x\) \(\text { I.F }=e^{\int \tan x d x}=-\sec x\) Solution is (I.F) \(=\int(\) I.F \() \sec x d x\) \(\frac{1}{y} \sec x=\tan x+c\)
AIEEE-2010
Differential Equation
87555
The curve amongst the family of curves represented by the differential equation, \(\left(x^{2}-\right.\) \(\left.\mathbf{y}^{2}\right) \mathbf{d x}+\mathbf{2 x y d y}=0\), which passes through \((1,1)\), is
1 a circle with centre on the Y-axis
2 a circle with centre on the X-axis
3 an ellipse with major axis along the \(\mathrm{Y}\)-axis
4 a hyperbola with transverse axis along the Xaxis.
Explanation:
(B) : \(\left(x^{2}-y^{2}\right) d x+2 x y d y=0\) Put, \(\quad y=v x\) \(\frac{d y}{d x}=\frac{y^{2}-x^{2}}{2 x y}\) \(y=v x\) \(\frac{d y}{d x}=v+x \frac{d v}{d x}\) Solving we get- \(\int \frac{2 v}{v^{2}+1} d v=\int-\frac{d x}{x}\) \(\ln \left(v^{2}+1\right)=-\ln x+C\) \(\mathrm{y}^{2}+\mathrm{x}^{2}=\mathrm{Cx}\) \(1+1=\mathrm{C}\) \(\mathrm{C}=2\) \(y^{2}+x^{2}=2 x\)
