87391
The differential equation for the family of curves \(x^{2}+y^{2}-2 a y=0\), where \(a\) is an arbitrary constant, is
1 \(2\left(x^{2}-y^{2}\right) y^{\prime}=x y\)
2 \(2\left(x^{2}+y^{2}\right) y^{\prime}=x y\)
3 \(\left(x^{2}-y^{2}\right) y^{\prime}=2 x y\)
4 \(\left(x^{2}+y^{2}\right) y^{\prime}=2 x y\)
Explanation:
(C) : Given the curves, \(x^{2}+y^{2}-2 a y=0\) \(2 x+2 y \cdot y^{\prime}-2 a y^{\prime}=0\) \(\therefore \quad \mathrm{a}=\frac{\left(\mathrm{x}+\mathrm{yy^{ \prime } )}\right.}{\mathrm{y}^{\prime}}\) Putting the value of a in equation (i), we get- \(\left(x^{2}+y^{2}\right)=\frac{2\left(x+y y^{\prime}\right) y}{y^{\prime}}\) or \(\quad\left(x^{2}+y^{2}\right) y^{\prime}=2 x y+2 y^{2} \cdot y^{\prime}\) or \(\quad\left(x^{2}-y^{2}\right) y^{\prime}=2 x y\)
AIEEE-2004
Differential Equation
87190
If \(\frac{d y}{d x}=e^{-2 y}\) and \(y=0\) when \(x=5\), then the value of \(x\) for \(y=3\) is
1 \(\mathrm{e}^{5}\)
2 \(\mathrm{e}^{6}+1\)
3 \(\frac{\mathrm{e}^{6}+9}{2}\)
4 none of these
Explanation:
(C) : Given differential equation, \(\frac{d y}{d x}=e^{-2 y} \Rightarrow e^{2 y} d y=d x\) Integrating both sides, ...(i) \(\int e^{2 y} d y=\int d x \Rightarrow \frac{e^{2 y}}{2}=x+c\) When, \(x=5, y=0\) \(\frac{\mathrm{e}^{0}}{2}=5+c \Rightarrow 1=10+2 \mathrm{c}\) \(2 \mathrm{c}=-9 \Rightarrow \mathrm{c}=-\frac{9}{2}\) Putting the value of ' \(c\) ' in equation (i), we get - \(\frac{\mathrm{e}^{2 \mathrm{y}}}{2}=\mathrm{x}-\frac{9}{2}\) When, \(y=3\), we get \(\frac{e^{6}}{2}=x-\frac{9}{2} \Rightarrow x=\frac{e^{6}+9}{2}\)
SRM JEEE-2010
Differential Equation
87191
The general solution of a differential equation \(\frac{d y}{d x}=\frac{x^{2}}{y^{2}}\) is
1 \(x_{2}^{3}-y^{3}=C\)
2 \(x^{3}+y^{3}=C\)
3 \(\mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{C}\)
4 \(\mathrm{x}^{2}-\mathrm{y}^{2}=\mathrm{C}\)
Explanation:
(A) : Given differential equation, \(\frac{d y}{d x} =\frac{x^{2}}{y^{2}}\) \(y^{2} d y =x^{2} d x\) Integrating both side we get, \(\frac{y^{3}}{3}=\frac{x^{3}}{3}+C_{1}\) \(y^{3}-x^{3}=3 C_{1}\) \(x^{3}-y^{3}=-3 C_{1} C=-3 C_{1}\) \(x^{3}-y^{3}=C\)
SRM JEEE-2009
Differential Equation
87201
\(y=a e^{\mathrm{mtx}}+b e^{-\mathrm{mlx}}\) satisfies which of the following differential equations :
1 \(\frac{d y}{d x}+m y=0\)
2 \(\frac{d y}{d x}-m y=0\)
3 \(\frac{d^{2} y}{d x^{2}}-m^{2} y=0\)
4 \(\frac{d^{2} y}{d x^{2}}+m^{2} y=0\)
Explanation:
(C) : We have, \(y=a e^{m x}+b e^{-m x}\) Differentiating with respect to \(\mathrm{x}\), we get- \(\frac{d y}{d x}=a^{m} e^{m x}-b^{-m x}\) Again differentiating with respect to \(x\), we get - \(\frac{d^{2} y}{d x^{2}}=a m^{2} e^{m x}+b m^{2} e^{-m x} \Rightarrow \frac{d^{2} y}{d x^{2}}=m^{2}\left(a e^{m x}+b e^{-m x}\right)\) \(\frac{d^{2} y}{d x^{2}}=m^{2} y\) \(\frac{d^{2} y}{d x^{2}}-m^{2} y=0\)
Karnataka CET-2002]**#
Differential Equation
87202
The general solution of the differential equation \(\frac{d y}{d x}+\frac{1+\cos 2 y}{1-\cos 2 x}=0\) is given by :
1 tany \(+\cos x=c\)
2 tany \(-\cot x=c\)
3 \(\tan x-\cot y=c\)
4 \(\tan x+\cot x=c\)
Explanation:
(B) : We have differential equation- \(\frac{d y}{d x}+\frac{1+\cos 2 y}{1-\cos 2 x}=0\) We know that, \(\cos 2 \theta=2 \cos ^{2} \theta-1 \Rightarrow \cos 2 \theta=1-2 \sin ^{2} \theta\) Then, \(\frac{d y}{d x}+\frac{2 \cos ^{2} y}{2 \sin ^{2} x}=0\) \(\frac{d y}{d x}=-\frac{\cos ^{2} y}{\sin ^{2} x} \Rightarrow \frac{d y}{\cos ^{2} y}+\frac{d x}{\sin ^{2} x}=0\) Integrating on both side, we get- \(\int \frac{d y}{\cos ^{2} y}+\int \frac{d x}{\sin ^{2} x}=0\) \(\int \sec ^{2} y d y+\int \operatorname{cosec}^{2} x=0\) \(\tan y-\cot x=c\)
87391
The differential equation for the family of curves \(x^{2}+y^{2}-2 a y=0\), where \(a\) is an arbitrary constant, is
1 \(2\left(x^{2}-y^{2}\right) y^{\prime}=x y\)
2 \(2\left(x^{2}+y^{2}\right) y^{\prime}=x y\)
3 \(\left(x^{2}-y^{2}\right) y^{\prime}=2 x y\)
4 \(\left(x^{2}+y^{2}\right) y^{\prime}=2 x y\)
Explanation:
(C) : Given the curves, \(x^{2}+y^{2}-2 a y=0\) \(2 x+2 y \cdot y^{\prime}-2 a y^{\prime}=0\) \(\therefore \quad \mathrm{a}=\frac{\left(\mathrm{x}+\mathrm{yy^{ \prime } )}\right.}{\mathrm{y}^{\prime}}\) Putting the value of a in equation (i), we get- \(\left(x^{2}+y^{2}\right)=\frac{2\left(x+y y^{\prime}\right) y}{y^{\prime}}\) or \(\quad\left(x^{2}+y^{2}\right) y^{\prime}=2 x y+2 y^{2} \cdot y^{\prime}\) or \(\quad\left(x^{2}-y^{2}\right) y^{\prime}=2 x y\)
AIEEE-2004
Differential Equation
87190
If \(\frac{d y}{d x}=e^{-2 y}\) and \(y=0\) when \(x=5\), then the value of \(x\) for \(y=3\) is
1 \(\mathrm{e}^{5}\)
2 \(\mathrm{e}^{6}+1\)
3 \(\frac{\mathrm{e}^{6}+9}{2}\)
4 none of these
Explanation:
(C) : Given differential equation, \(\frac{d y}{d x}=e^{-2 y} \Rightarrow e^{2 y} d y=d x\) Integrating both sides, ...(i) \(\int e^{2 y} d y=\int d x \Rightarrow \frac{e^{2 y}}{2}=x+c\) When, \(x=5, y=0\) \(\frac{\mathrm{e}^{0}}{2}=5+c \Rightarrow 1=10+2 \mathrm{c}\) \(2 \mathrm{c}=-9 \Rightarrow \mathrm{c}=-\frac{9}{2}\) Putting the value of ' \(c\) ' in equation (i), we get - \(\frac{\mathrm{e}^{2 \mathrm{y}}}{2}=\mathrm{x}-\frac{9}{2}\) When, \(y=3\), we get \(\frac{e^{6}}{2}=x-\frac{9}{2} \Rightarrow x=\frac{e^{6}+9}{2}\)
SRM JEEE-2010
Differential Equation
87191
The general solution of a differential equation \(\frac{d y}{d x}=\frac{x^{2}}{y^{2}}\) is
1 \(x_{2}^{3}-y^{3}=C\)
2 \(x^{3}+y^{3}=C\)
3 \(\mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{C}\)
4 \(\mathrm{x}^{2}-\mathrm{y}^{2}=\mathrm{C}\)
Explanation:
(A) : Given differential equation, \(\frac{d y}{d x} =\frac{x^{2}}{y^{2}}\) \(y^{2} d y =x^{2} d x\) Integrating both side we get, \(\frac{y^{3}}{3}=\frac{x^{3}}{3}+C_{1}\) \(y^{3}-x^{3}=3 C_{1}\) \(x^{3}-y^{3}=-3 C_{1} C=-3 C_{1}\) \(x^{3}-y^{3}=C\)
SRM JEEE-2009
Differential Equation
87201
\(y=a e^{\mathrm{mtx}}+b e^{-\mathrm{mlx}}\) satisfies which of the following differential equations :
1 \(\frac{d y}{d x}+m y=0\)
2 \(\frac{d y}{d x}-m y=0\)
3 \(\frac{d^{2} y}{d x^{2}}-m^{2} y=0\)
4 \(\frac{d^{2} y}{d x^{2}}+m^{2} y=0\)
Explanation:
(C) : We have, \(y=a e^{m x}+b e^{-m x}\) Differentiating with respect to \(\mathrm{x}\), we get- \(\frac{d y}{d x}=a^{m} e^{m x}-b^{-m x}\) Again differentiating with respect to \(x\), we get - \(\frac{d^{2} y}{d x^{2}}=a m^{2} e^{m x}+b m^{2} e^{-m x} \Rightarrow \frac{d^{2} y}{d x^{2}}=m^{2}\left(a e^{m x}+b e^{-m x}\right)\) \(\frac{d^{2} y}{d x^{2}}=m^{2} y\) \(\frac{d^{2} y}{d x^{2}}-m^{2} y=0\)
Karnataka CET-2002]**#
Differential Equation
87202
The general solution of the differential equation \(\frac{d y}{d x}+\frac{1+\cos 2 y}{1-\cos 2 x}=0\) is given by :
1 tany \(+\cos x=c\)
2 tany \(-\cot x=c\)
3 \(\tan x-\cot y=c\)
4 \(\tan x+\cot x=c\)
Explanation:
(B) : We have differential equation- \(\frac{d y}{d x}+\frac{1+\cos 2 y}{1-\cos 2 x}=0\) We know that, \(\cos 2 \theta=2 \cos ^{2} \theta-1 \Rightarrow \cos 2 \theta=1-2 \sin ^{2} \theta\) Then, \(\frac{d y}{d x}+\frac{2 \cos ^{2} y}{2 \sin ^{2} x}=0\) \(\frac{d y}{d x}=-\frac{\cos ^{2} y}{\sin ^{2} x} \Rightarrow \frac{d y}{\cos ^{2} y}+\frac{d x}{\sin ^{2} x}=0\) Integrating on both side, we get- \(\int \frac{d y}{\cos ^{2} y}+\int \frac{d x}{\sin ^{2} x}=0\) \(\int \sec ^{2} y d y+\int \operatorname{cosec}^{2} x=0\) \(\tan y-\cot x=c\)
87391
The differential equation for the family of curves \(x^{2}+y^{2}-2 a y=0\), where \(a\) is an arbitrary constant, is
1 \(2\left(x^{2}-y^{2}\right) y^{\prime}=x y\)
2 \(2\left(x^{2}+y^{2}\right) y^{\prime}=x y\)
3 \(\left(x^{2}-y^{2}\right) y^{\prime}=2 x y\)
4 \(\left(x^{2}+y^{2}\right) y^{\prime}=2 x y\)
Explanation:
(C) : Given the curves, \(x^{2}+y^{2}-2 a y=0\) \(2 x+2 y \cdot y^{\prime}-2 a y^{\prime}=0\) \(\therefore \quad \mathrm{a}=\frac{\left(\mathrm{x}+\mathrm{yy^{ \prime } )}\right.}{\mathrm{y}^{\prime}}\) Putting the value of a in equation (i), we get- \(\left(x^{2}+y^{2}\right)=\frac{2\left(x+y y^{\prime}\right) y}{y^{\prime}}\) or \(\quad\left(x^{2}+y^{2}\right) y^{\prime}=2 x y+2 y^{2} \cdot y^{\prime}\) or \(\quad\left(x^{2}-y^{2}\right) y^{\prime}=2 x y\)
AIEEE-2004
Differential Equation
87190
If \(\frac{d y}{d x}=e^{-2 y}\) and \(y=0\) when \(x=5\), then the value of \(x\) for \(y=3\) is
1 \(\mathrm{e}^{5}\)
2 \(\mathrm{e}^{6}+1\)
3 \(\frac{\mathrm{e}^{6}+9}{2}\)
4 none of these
Explanation:
(C) : Given differential equation, \(\frac{d y}{d x}=e^{-2 y} \Rightarrow e^{2 y} d y=d x\) Integrating both sides, ...(i) \(\int e^{2 y} d y=\int d x \Rightarrow \frac{e^{2 y}}{2}=x+c\) When, \(x=5, y=0\) \(\frac{\mathrm{e}^{0}}{2}=5+c \Rightarrow 1=10+2 \mathrm{c}\) \(2 \mathrm{c}=-9 \Rightarrow \mathrm{c}=-\frac{9}{2}\) Putting the value of ' \(c\) ' in equation (i), we get - \(\frac{\mathrm{e}^{2 \mathrm{y}}}{2}=\mathrm{x}-\frac{9}{2}\) When, \(y=3\), we get \(\frac{e^{6}}{2}=x-\frac{9}{2} \Rightarrow x=\frac{e^{6}+9}{2}\)
SRM JEEE-2010
Differential Equation
87191
The general solution of a differential equation \(\frac{d y}{d x}=\frac{x^{2}}{y^{2}}\) is
1 \(x_{2}^{3}-y^{3}=C\)
2 \(x^{3}+y^{3}=C\)
3 \(\mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{C}\)
4 \(\mathrm{x}^{2}-\mathrm{y}^{2}=\mathrm{C}\)
Explanation:
(A) : Given differential equation, \(\frac{d y}{d x} =\frac{x^{2}}{y^{2}}\) \(y^{2} d y =x^{2} d x\) Integrating both side we get, \(\frac{y^{3}}{3}=\frac{x^{3}}{3}+C_{1}\) \(y^{3}-x^{3}=3 C_{1}\) \(x^{3}-y^{3}=-3 C_{1} C=-3 C_{1}\) \(x^{3}-y^{3}=C\)
SRM JEEE-2009
Differential Equation
87201
\(y=a e^{\mathrm{mtx}}+b e^{-\mathrm{mlx}}\) satisfies which of the following differential equations :
1 \(\frac{d y}{d x}+m y=0\)
2 \(\frac{d y}{d x}-m y=0\)
3 \(\frac{d^{2} y}{d x^{2}}-m^{2} y=0\)
4 \(\frac{d^{2} y}{d x^{2}}+m^{2} y=0\)
Explanation:
(C) : We have, \(y=a e^{m x}+b e^{-m x}\) Differentiating with respect to \(\mathrm{x}\), we get- \(\frac{d y}{d x}=a^{m} e^{m x}-b^{-m x}\) Again differentiating with respect to \(x\), we get - \(\frac{d^{2} y}{d x^{2}}=a m^{2} e^{m x}+b m^{2} e^{-m x} \Rightarrow \frac{d^{2} y}{d x^{2}}=m^{2}\left(a e^{m x}+b e^{-m x}\right)\) \(\frac{d^{2} y}{d x^{2}}=m^{2} y\) \(\frac{d^{2} y}{d x^{2}}-m^{2} y=0\)
Karnataka CET-2002]**#
Differential Equation
87202
The general solution of the differential equation \(\frac{d y}{d x}+\frac{1+\cos 2 y}{1-\cos 2 x}=0\) is given by :
1 tany \(+\cos x=c\)
2 tany \(-\cot x=c\)
3 \(\tan x-\cot y=c\)
4 \(\tan x+\cot x=c\)
Explanation:
(B) : We have differential equation- \(\frac{d y}{d x}+\frac{1+\cos 2 y}{1-\cos 2 x}=0\) We know that, \(\cos 2 \theta=2 \cos ^{2} \theta-1 \Rightarrow \cos 2 \theta=1-2 \sin ^{2} \theta\) Then, \(\frac{d y}{d x}+\frac{2 \cos ^{2} y}{2 \sin ^{2} x}=0\) \(\frac{d y}{d x}=-\frac{\cos ^{2} y}{\sin ^{2} x} \Rightarrow \frac{d y}{\cos ^{2} y}+\frac{d x}{\sin ^{2} x}=0\) Integrating on both side, we get- \(\int \frac{d y}{\cos ^{2} y}+\int \frac{d x}{\sin ^{2} x}=0\) \(\int \sec ^{2} y d y+\int \operatorname{cosec}^{2} x=0\) \(\tan y-\cot x=c\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Differential Equation
87391
The differential equation for the family of curves \(x^{2}+y^{2}-2 a y=0\), where \(a\) is an arbitrary constant, is
1 \(2\left(x^{2}-y^{2}\right) y^{\prime}=x y\)
2 \(2\left(x^{2}+y^{2}\right) y^{\prime}=x y\)
3 \(\left(x^{2}-y^{2}\right) y^{\prime}=2 x y\)
4 \(\left(x^{2}+y^{2}\right) y^{\prime}=2 x y\)
Explanation:
(C) : Given the curves, \(x^{2}+y^{2}-2 a y=0\) \(2 x+2 y \cdot y^{\prime}-2 a y^{\prime}=0\) \(\therefore \quad \mathrm{a}=\frac{\left(\mathrm{x}+\mathrm{yy^{ \prime } )}\right.}{\mathrm{y}^{\prime}}\) Putting the value of a in equation (i), we get- \(\left(x^{2}+y^{2}\right)=\frac{2\left(x+y y^{\prime}\right) y}{y^{\prime}}\) or \(\quad\left(x^{2}+y^{2}\right) y^{\prime}=2 x y+2 y^{2} \cdot y^{\prime}\) or \(\quad\left(x^{2}-y^{2}\right) y^{\prime}=2 x y\)
AIEEE-2004
Differential Equation
87190
If \(\frac{d y}{d x}=e^{-2 y}\) and \(y=0\) when \(x=5\), then the value of \(x\) for \(y=3\) is
1 \(\mathrm{e}^{5}\)
2 \(\mathrm{e}^{6}+1\)
3 \(\frac{\mathrm{e}^{6}+9}{2}\)
4 none of these
Explanation:
(C) : Given differential equation, \(\frac{d y}{d x}=e^{-2 y} \Rightarrow e^{2 y} d y=d x\) Integrating both sides, ...(i) \(\int e^{2 y} d y=\int d x \Rightarrow \frac{e^{2 y}}{2}=x+c\) When, \(x=5, y=0\) \(\frac{\mathrm{e}^{0}}{2}=5+c \Rightarrow 1=10+2 \mathrm{c}\) \(2 \mathrm{c}=-9 \Rightarrow \mathrm{c}=-\frac{9}{2}\) Putting the value of ' \(c\) ' in equation (i), we get - \(\frac{\mathrm{e}^{2 \mathrm{y}}}{2}=\mathrm{x}-\frac{9}{2}\) When, \(y=3\), we get \(\frac{e^{6}}{2}=x-\frac{9}{2} \Rightarrow x=\frac{e^{6}+9}{2}\)
SRM JEEE-2010
Differential Equation
87191
The general solution of a differential equation \(\frac{d y}{d x}=\frac{x^{2}}{y^{2}}\) is
1 \(x_{2}^{3}-y^{3}=C\)
2 \(x^{3}+y^{3}=C\)
3 \(\mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{C}\)
4 \(\mathrm{x}^{2}-\mathrm{y}^{2}=\mathrm{C}\)
Explanation:
(A) : Given differential equation, \(\frac{d y}{d x} =\frac{x^{2}}{y^{2}}\) \(y^{2} d y =x^{2} d x\) Integrating both side we get, \(\frac{y^{3}}{3}=\frac{x^{3}}{3}+C_{1}\) \(y^{3}-x^{3}=3 C_{1}\) \(x^{3}-y^{3}=-3 C_{1} C=-3 C_{1}\) \(x^{3}-y^{3}=C\)
SRM JEEE-2009
Differential Equation
87201
\(y=a e^{\mathrm{mtx}}+b e^{-\mathrm{mlx}}\) satisfies which of the following differential equations :
1 \(\frac{d y}{d x}+m y=0\)
2 \(\frac{d y}{d x}-m y=0\)
3 \(\frac{d^{2} y}{d x^{2}}-m^{2} y=0\)
4 \(\frac{d^{2} y}{d x^{2}}+m^{2} y=0\)
Explanation:
(C) : We have, \(y=a e^{m x}+b e^{-m x}\) Differentiating with respect to \(\mathrm{x}\), we get- \(\frac{d y}{d x}=a^{m} e^{m x}-b^{-m x}\) Again differentiating with respect to \(x\), we get - \(\frac{d^{2} y}{d x^{2}}=a m^{2} e^{m x}+b m^{2} e^{-m x} \Rightarrow \frac{d^{2} y}{d x^{2}}=m^{2}\left(a e^{m x}+b e^{-m x}\right)\) \(\frac{d^{2} y}{d x^{2}}=m^{2} y\) \(\frac{d^{2} y}{d x^{2}}-m^{2} y=0\)
Karnataka CET-2002]**#
Differential Equation
87202
The general solution of the differential equation \(\frac{d y}{d x}+\frac{1+\cos 2 y}{1-\cos 2 x}=0\) is given by :
1 tany \(+\cos x=c\)
2 tany \(-\cot x=c\)
3 \(\tan x-\cot y=c\)
4 \(\tan x+\cot x=c\)
Explanation:
(B) : We have differential equation- \(\frac{d y}{d x}+\frac{1+\cos 2 y}{1-\cos 2 x}=0\) We know that, \(\cos 2 \theta=2 \cos ^{2} \theta-1 \Rightarrow \cos 2 \theta=1-2 \sin ^{2} \theta\) Then, \(\frac{d y}{d x}+\frac{2 \cos ^{2} y}{2 \sin ^{2} x}=0\) \(\frac{d y}{d x}=-\frac{\cos ^{2} y}{\sin ^{2} x} \Rightarrow \frac{d y}{\cos ^{2} y}+\frac{d x}{\sin ^{2} x}=0\) Integrating on both side, we get- \(\int \frac{d y}{\cos ^{2} y}+\int \frac{d x}{\sin ^{2} x}=0\) \(\int \sec ^{2} y d y+\int \operatorname{cosec}^{2} x=0\) \(\tan y-\cot x=c\)
87391
The differential equation for the family of curves \(x^{2}+y^{2}-2 a y=0\), where \(a\) is an arbitrary constant, is
1 \(2\left(x^{2}-y^{2}\right) y^{\prime}=x y\)
2 \(2\left(x^{2}+y^{2}\right) y^{\prime}=x y\)
3 \(\left(x^{2}-y^{2}\right) y^{\prime}=2 x y\)
4 \(\left(x^{2}+y^{2}\right) y^{\prime}=2 x y\)
Explanation:
(C) : Given the curves, \(x^{2}+y^{2}-2 a y=0\) \(2 x+2 y \cdot y^{\prime}-2 a y^{\prime}=0\) \(\therefore \quad \mathrm{a}=\frac{\left(\mathrm{x}+\mathrm{yy^{ \prime } )}\right.}{\mathrm{y}^{\prime}}\) Putting the value of a in equation (i), we get- \(\left(x^{2}+y^{2}\right)=\frac{2\left(x+y y^{\prime}\right) y}{y^{\prime}}\) or \(\quad\left(x^{2}+y^{2}\right) y^{\prime}=2 x y+2 y^{2} \cdot y^{\prime}\) or \(\quad\left(x^{2}-y^{2}\right) y^{\prime}=2 x y\)
AIEEE-2004
Differential Equation
87190
If \(\frac{d y}{d x}=e^{-2 y}\) and \(y=0\) when \(x=5\), then the value of \(x\) for \(y=3\) is
1 \(\mathrm{e}^{5}\)
2 \(\mathrm{e}^{6}+1\)
3 \(\frac{\mathrm{e}^{6}+9}{2}\)
4 none of these
Explanation:
(C) : Given differential equation, \(\frac{d y}{d x}=e^{-2 y} \Rightarrow e^{2 y} d y=d x\) Integrating both sides, ...(i) \(\int e^{2 y} d y=\int d x \Rightarrow \frac{e^{2 y}}{2}=x+c\) When, \(x=5, y=0\) \(\frac{\mathrm{e}^{0}}{2}=5+c \Rightarrow 1=10+2 \mathrm{c}\) \(2 \mathrm{c}=-9 \Rightarrow \mathrm{c}=-\frac{9}{2}\) Putting the value of ' \(c\) ' in equation (i), we get - \(\frac{\mathrm{e}^{2 \mathrm{y}}}{2}=\mathrm{x}-\frac{9}{2}\) When, \(y=3\), we get \(\frac{e^{6}}{2}=x-\frac{9}{2} \Rightarrow x=\frac{e^{6}+9}{2}\)
SRM JEEE-2010
Differential Equation
87191
The general solution of a differential equation \(\frac{d y}{d x}=\frac{x^{2}}{y^{2}}\) is
1 \(x_{2}^{3}-y^{3}=C\)
2 \(x^{3}+y^{3}=C\)
3 \(\mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{C}\)
4 \(\mathrm{x}^{2}-\mathrm{y}^{2}=\mathrm{C}\)
Explanation:
(A) : Given differential equation, \(\frac{d y}{d x} =\frac{x^{2}}{y^{2}}\) \(y^{2} d y =x^{2} d x\) Integrating both side we get, \(\frac{y^{3}}{3}=\frac{x^{3}}{3}+C_{1}\) \(y^{3}-x^{3}=3 C_{1}\) \(x^{3}-y^{3}=-3 C_{1} C=-3 C_{1}\) \(x^{3}-y^{3}=C\)
SRM JEEE-2009
Differential Equation
87201
\(y=a e^{\mathrm{mtx}}+b e^{-\mathrm{mlx}}\) satisfies which of the following differential equations :
1 \(\frac{d y}{d x}+m y=0\)
2 \(\frac{d y}{d x}-m y=0\)
3 \(\frac{d^{2} y}{d x^{2}}-m^{2} y=0\)
4 \(\frac{d^{2} y}{d x^{2}}+m^{2} y=0\)
Explanation:
(C) : We have, \(y=a e^{m x}+b e^{-m x}\) Differentiating with respect to \(\mathrm{x}\), we get- \(\frac{d y}{d x}=a^{m} e^{m x}-b^{-m x}\) Again differentiating with respect to \(x\), we get - \(\frac{d^{2} y}{d x^{2}}=a m^{2} e^{m x}+b m^{2} e^{-m x} \Rightarrow \frac{d^{2} y}{d x^{2}}=m^{2}\left(a e^{m x}+b e^{-m x}\right)\) \(\frac{d^{2} y}{d x^{2}}=m^{2} y\) \(\frac{d^{2} y}{d x^{2}}-m^{2} y=0\)
Karnataka CET-2002]**#
Differential Equation
87202
The general solution of the differential equation \(\frac{d y}{d x}+\frac{1+\cos 2 y}{1-\cos 2 x}=0\) is given by :
1 tany \(+\cos x=c\)
2 tany \(-\cot x=c\)
3 \(\tan x-\cot y=c\)
4 \(\tan x+\cot x=c\)
Explanation:
(B) : We have differential equation- \(\frac{d y}{d x}+\frac{1+\cos 2 y}{1-\cos 2 x}=0\) We know that, \(\cos 2 \theta=2 \cos ^{2} \theta-1 \Rightarrow \cos 2 \theta=1-2 \sin ^{2} \theta\) Then, \(\frac{d y}{d x}+\frac{2 \cos ^{2} y}{2 \sin ^{2} x}=0\) \(\frac{d y}{d x}=-\frac{\cos ^{2} y}{\sin ^{2} x} \Rightarrow \frac{d y}{\cos ^{2} y}+\frac{d x}{\sin ^{2} x}=0\) Integrating on both side, we get- \(\int \frac{d y}{\cos ^{2} y}+\int \frac{d x}{\sin ^{2} x}=0\) \(\int \sec ^{2} y d y+\int \operatorname{cosec}^{2} x=0\) \(\tan y-\cot x=c\)