87259
The differential equation representing the family of parabolas having vertex at origin and axis along positive direction of \(x\)-axis is
1 \(y^{2}-2 x y \frac{d y}{d x}=0\)
2 \(y^{2}+2 x y \frac{d y}{d x}=0\)
3 \(y^{2}-2 x y \frac{d^{2} y}{d x^{2}}=0\)
4 \(y^{2}+2 x y \frac{d^{2} y}{d x^{2}}=0\)
Explanation:
(A) : Differential equation representing the family of parabola having vertex at origin is \(y^{2}=4 a x \tag{i}\) Differentiating equation (i) w.r.t.x, we get - \(2 y \frac{d y}{d x}=4 a \Rightarrow 2 y \frac{d y}{d x}=4 \times \frac{y^{2}}{4 x} \quad[\because\) from equation \((i)]\) \(2 y \frac{d y}{d x}=\frac{y^{2}}{x} \Rightarrow 2 y x \frac{d y}{d x}=y^{2}\) \(\mathrm{y}^{2}-2 \mathrm{yx} \frac{\mathrm{dy}}{\mathrm{dx}}=0\)
AMU-2010
Differential Equation
87260
The differential equation representing the family of curves \(y=b \sin (x+a)\), where \(a, b\) are arbitrary constants is
1 \(\frac{d^{2} y}{d x^{2}}+y=0\)
2 \(\frac{d^{2} y}{d x^{2}}-y=0\)
3 \(\frac{d y}{d x}+y=0\)
4 none of these
Explanation:
(A) : The given family of curve is, \(y=b \sin (x+a)\) Differentiating (i) with respect to \(\mathrm{x}\), we get \(\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{b} \cos (\mathrm{x}+\mathrm{a})\) Differentiating (ii) with respect to \(x\), we get \(\frac{d^{2} y}{d x^{2}}=-b \sin (x+a) \Rightarrow \frac{d^{2} y}{d x^{2}}=-y\) \(\frac{d^{2} y}{d x^{2}}+y=0\)
AMU-2010
Differential Equation
87261
The general solution of the differential equation \(y d x+\left(x+2 y^{2}\right) d y=0\) is
1 \(x y+y^{2}=c\)
2 \(3 x y+y^{2}=c\)
3 \(x y+y^{3}=c\)
4 \(3 x y+2 y^{3}=c\)
Explanation:
(D) : The given differential equation is, \(y d x+\left(x+2 y^{2}\right) d y=0\) \(y d x+x d y+2 y^{2} d y=0\) \(d(x y)+2 y^{2} d y=0\) Integrating both sides, we get - \(\mathrm{xy}+\frac{2 \mathrm{y}^{3}}{3}=\) constant \(3 x y+2 y^{3}=C\)
87259
The differential equation representing the family of parabolas having vertex at origin and axis along positive direction of \(x\)-axis is
1 \(y^{2}-2 x y \frac{d y}{d x}=0\)
2 \(y^{2}+2 x y \frac{d y}{d x}=0\)
3 \(y^{2}-2 x y \frac{d^{2} y}{d x^{2}}=0\)
4 \(y^{2}+2 x y \frac{d^{2} y}{d x^{2}}=0\)
Explanation:
(A) : Differential equation representing the family of parabola having vertex at origin is \(y^{2}=4 a x \tag{i}\) Differentiating equation (i) w.r.t.x, we get - \(2 y \frac{d y}{d x}=4 a \Rightarrow 2 y \frac{d y}{d x}=4 \times \frac{y^{2}}{4 x} \quad[\because\) from equation \((i)]\) \(2 y \frac{d y}{d x}=\frac{y^{2}}{x} \Rightarrow 2 y x \frac{d y}{d x}=y^{2}\) \(\mathrm{y}^{2}-2 \mathrm{yx} \frac{\mathrm{dy}}{\mathrm{dx}}=0\)
AMU-2010
Differential Equation
87260
The differential equation representing the family of curves \(y=b \sin (x+a)\), where \(a, b\) are arbitrary constants is
1 \(\frac{d^{2} y}{d x^{2}}+y=0\)
2 \(\frac{d^{2} y}{d x^{2}}-y=0\)
3 \(\frac{d y}{d x}+y=0\)
4 none of these
Explanation:
(A) : The given family of curve is, \(y=b \sin (x+a)\) Differentiating (i) with respect to \(\mathrm{x}\), we get \(\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{b} \cos (\mathrm{x}+\mathrm{a})\) Differentiating (ii) with respect to \(x\), we get \(\frac{d^{2} y}{d x^{2}}=-b \sin (x+a) \Rightarrow \frac{d^{2} y}{d x^{2}}=-y\) \(\frac{d^{2} y}{d x^{2}}+y=0\)
AMU-2010
Differential Equation
87261
The general solution of the differential equation \(y d x+\left(x+2 y^{2}\right) d y=0\) is
1 \(x y+y^{2}=c\)
2 \(3 x y+y^{2}=c\)
3 \(x y+y^{3}=c\)
4 \(3 x y+2 y^{3}=c\)
Explanation:
(D) : The given differential equation is, \(y d x+\left(x+2 y^{2}\right) d y=0\) \(y d x+x d y+2 y^{2} d y=0\) \(d(x y)+2 y^{2} d y=0\) Integrating both sides, we get - \(\mathrm{xy}+\frac{2 \mathrm{y}^{3}}{3}=\) constant \(3 x y+2 y^{3}=C\)
87259
The differential equation representing the family of parabolas having vertex at origin and axis along positive direction of \(x\)-axis is
1 \(y^{2}-2 x y \frac{d y}{d x}=0\)
2 \(y^{2}+2 x y \frac{d y}{d x}=0\)
3 \(y^{2}-2 x y \frac{d^{2} y}{d x^{2}}=0\)
4 \(y^{2}+2 x y \frac{d^{2} y}{d x^{2}}=0\)
Explanation:
(A) : Differential equation representing the family of parabola having vertex at origin is \(y^{2}=4 a x \tag{i}\) Differentiating equation (i) w.r.t.x, we get - \(2 y \frac{d y}{d x}=4 a \Rightarrow 2 y \frac{d y}{d x}=4 \times \frac{y^{2}}{4 x} \quad[\because\) from equation \((i)]\) \(2 y \frac{d y}{d x}=\frac{y^{2}}{x} \Rightarrow 2 y x \frac{d y}{d x}=y^{2}\) \(\mathrm{y}^{2}-2 \mathrm{yx} \frac{\mathrm{dy}}{\mathrm{dx}}=0\)
AMU-2010
Differential Equation
87260
The differential equation representing the family of curves \(y=b \sin (x+a)\), where \(a, b\) are arbitrary constants is
1 \(\frac{d^{2} y}{d x^{2}}+y=0\)
2 \(\frac{d^{2} y}{d x^{2}}-y=0\)
3 \(\frac{d y}{d x}+y=0\)
4 none of these
Explanation:
(A) : The given family of curve is, \(y=b \sin (x+a)\) Differentiating (i) with respect to \(\mathrm{x}\), we get \(\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{b} \cos (\mathrm{x}+\mathrm{a})\) Differentiating (ii) with respect to \(x\), we get \(\frac{d^{2} y}{d x^{2}}=-b \sin (x+a) \Rightarrow \frac{d^{2} y}{d x^{2}}=-y\) \(\frac{d^{2} y}{d x^{2}}+y=0\)
AMU-2010
Differential Equation
87261
The general solution of the differential equation \(y d x+\left(x+2 y^{2}\right) d y=0\) is
1 \(x y+y^{2}=c\)
2 \(3 x y+y^{2}=c\)
3 \(x y+y^{3}=c\)
4 \(3 x y+2 y^{3}=c\)
Explanation:
(D) : The given differential equation is, \(y d x+\left(x+2 y^{2}\right) d y=0\) \(y d x+x d y+2 y^{2} d y=0\) \(d(x y)+2 y^{2} d y=0\) Integrating both sides, we get - \(\mathrm{xy}+\frac{2 \mathrm{y}^{3}}{3}=\) constant \(3 x y+2 y^{3}=C\)
87259
The differential equation representing the family of parabolas having vertex at origin and axis along positive direction of \(x\)-axis is
1 \(y^{2}-2 x y \frac{d y}{d x}=0\)
2 \(y^{2}+2 x y \frac{d y}{d x}=0\)
3 \(y^{2}-2 x y \frac{d^{2} y}{d x^{2}}=0\)
4 \(y^{2}+2 x y \frac{d^{2} y}{d x^{2}}=0\)
Explanation:
(A) : Differential equation representing the family of parabola having vertex at origin is \(y^{2}=4 a x \tag{i}\) Differentiating equation (i) w.r.t.x, we get - \(2 y \frac{d y}{d x}=4 a \Rightarrow 2 y \frac{d y}{d x}=4 \times \frac{y^{2}}{4 x} \quad[\because\) from equation \((i)]\) \(2 y \frac{d y}{d x}=\frac{y^{2}}{x} \Rightarrow 2 y x \frac{d y}{d x}=y^{2}\) \(\mathrm{y}^{2}-2 \mathrm{yx} \frac{\mathrm{dy}}{\mathrm{dx}}=0\)
AMU-2010
Differential Equation
87260
The differential equation representing the family of curves \(y=b \sin (x+a)\), where \(a, b\) are arbitrary constants is
1 \(\frac{d^{2} y}{d x^{2}}+y=0\)
2 \(\frac{d^{2} y}{d x^{2}}-y=0\)
3 \(\frac{d y}{d x}+y=0\)
4 none of these
Explanation:
(A) : The given family of curve is, \(y=b \sin (x+a)\) Differentiating (i) with respect to \(\mathrm{x}\), we get \(\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{b} \cos (\mathrm{x}+\mathrm{a})\) Differentiating (ii) with respect to \(x\), we get \(\frac{d^{2} y}{d x^{2}}=-b \sin (x+a) \Rightarrow \frac{d^{2} y}{d x^{2}}=-y\) \(\frac{d^{2} y}{d x^{2}}+y=0\)
AMU-2010
Differential Equation
87261
The general solution of the differential equation \(y d x+\left(x+2 y^{2}\right) d y=0\) is
1 \(x y+y^{2}=c\)
2 \(3 x y+y^{2}=c\)
3 \(x y+y^{3}=c\)
4 \(3 x y+2 y^{3}=c\)
Explanation:
(D) : The given differential equation is, \(y d x+\left(x+2 y^{2}\right) d y=0\) \(y d x+x d y+2 y^{2} d y=0\) \(d(x y)+2 y^{2} d y=0\) Integrating both sides, we get - \(\mathrm{xy}+\frac{2 \mathrm{y}^{3}}{3}=\) constant \(3 x y+2 y^{3}=C\)
87259
The differential equation representing the family of parabolas having vertex at origin and axis along positive direction of \(x\)-axis is
1 \(y^{2}-2 x y \frac{d y}{d x}=0\)
2 \(y^{2}+2 x y \frac{d y}{d x}=0\)
3 \(y^{2}-2 x y \frac{d^{2} y}{d x^{2}}=0\)
4 \(y^{2}+2 x y \frac{d^{2} y}{d x^{2}}=0\)
Explanation:
(A) : Differential equation representing the family of parabola having vertex at origin is \(y^{2}=4 a x \tag{i}\) Differentiating equation (i) w.r.t.x, we get - \(2 y \frac{d y}{d x}=4 a \Rightarrow 2 y \frac{d y}{d x}=4 \times \frac{y^{2}}{4 x} \quad[\because\) from equation \((i)]\) \(2 y \frac{d y}{d x}=\frac{y^{2}}{x} \Rightarrow 2 y x \frac{d y}{d x}=y^{2}\) \(\mathrm{y}^{2}-2 \mathrm{yx} \frac{\mathrm{dy}}{\mathrm{dx}}=0\)
AMU-2010
Differential Equation
87260
The differential equation representing the family of curves \(y=b \sin (x+a)\), where \(a, b\) are arbitrary constants is
1 \(\frac{d^{2} y}{d x^{2}}+y=0\)
2 \(\frac{d^{2} y}{d x^{2}}-y=0\)
3 \(\frac{d y}{d x}+y=0\)
4 none of these
Explanation:
(A) : The given family of curve is, \(y=b \sin (x+a)\) Differentiating (i) with respect to \(\mathrm{x}\), we get \(\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{b} \cos (\mathrm{x}+\mathrm{a})\) Differentiating (ii) with respect to \(x\), we get \(\frac{d^{2} y}{d x^{2}}=-b \sin (x+a) \Rightarrow \frac{d^{2} y}{d x^{2}}=-y\) \(\frac{d^{2} y}{d x^{2}}+y=0\)
AMU-2010
Differential Equation
87261
The general solution of the differential equation \(y d x+\left(x+2 y^{2}\right) d y=0\) is
1 \(x y+y^{2}=c\)
2 \(3 x y+y^{2}=c\)
3 \(x y+y^{3}=c\)
4 \(3 x y+2 y^{3}=c\)
Explanation:
(D) : The given differential equation is, \(y d x+\left(x+2 y^{2}\right) d y=0\) \(y d x+x d y+2 y^{2} d y=0\) \(d(x y)+2 y^{2} d y=0\) Integrating both sides, we get - \(\mathrm{xy}+\frac{2 \mathrm{y}^{3}}{3}=\) constant \(3 x y+2 y^{3}=C\)
