87132
The differential equation whose solution is \(\mathbf{A x}\) \(+B y^{2}=1\), where \(A\) and \(B\) are arbitrary constant, is of
1 first order and second degree
2 first order and first degree
3 second order and first degree
4 second order and second degree
Explanation:
(C) : Given the solution, \(a x^{2}+b y^{2}=1\) Now differentiating equation (i) w.r.t. \(x\), we get- \(2 a x+2 b y \frac{d y}{d x}=0 \text { or } \frac{y}{x}\left(\frac{d y}{d x}\right)=-\frac{a}{b}\) Again differentiating w.r.t. \(\mathrm{x}\), we get \(\frac{x\left\{\left(\frac{d y}{d x}\right)^{2}+y\left(\frac{d^{2} y}{d x^{2}}\right)-y\left(\frac{d y}{d x}\right)\right\}}{x^{2}}=0\) or, \(\quad x y\left(\frac{d^{2} y}{d x^{2}}\right)+x\left(\frac{d y}{d x}\right)^{2}-y\left(\frac{d y}{d x}\right)=0\) This is the differential equation of order 2 and degree 1.
AIEEE-2006
Differential Equation
87133
The differential equation representing the family of curves \(y^{2}=2 c(x+\sqrt{c})\), where \(c>0\), is a parameter, is of order and degree as follows
1 order 2, degree 2
2 order 1, degree 3
3 order 1, degree 1
4 order1, degree 2
Explanation:
(B) : \(\mathrm{y}^{2}=2 \mathrm{c}(\mathrm{x}+\sqrt{\mathrm{c}})\) \(y^{2}=2 c x+2 c^{\frac{3}{2}}\) On differentiating with respect to \(x\), we get- \(2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{c} \Rightarrow \mathrm{c}=\mathrm{y}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)\) On substitution equation (ii) in (i) we get- \(y^{2}=2 x y\left(\frac{d y}{d x}\right)+2\left(y \frac{d y}{d x}\right)^{\frac{3}{2}}\) \(\left\{y^{2}-2 x y\left(\frac{d y}{d x}\right)\right\}^{2}=2^{2} y^{3}\left(\frac{d y}{d x}\right)^{3}\) \(\left\{y^{2}-2 x y\left(\frac{d y}{d x}\right)\right\}^{2}=4 y^{3}\left(\frac{d y}{d x}\right)^{3}\) order \(=\) highest differential equation Degree \(=\) highest degree of order \(\therefore\) order \(=1\) and degree \(=3\)
Jamia Millia Cslamia-2012
Differential Equation
87134
The degree of the differential equation \(y_{2}^{3 / 2}-y_{1}^{1 / 2}-4=0\) is
1 6
2 3
3 2
4 4
Explanation:
(A) : \(\mathrm{y}_{2}^{3 / 2}=\mathrm{y}_{1}^{1 / 2}+4\) Squaring on both sides we get \(y_{2}^{3}=y_{1}+16+8 y_{1}^{1 / 2}\) \(\left(y_{2}^{3}-y_{1}-16\right)^{2}=64 y_{1}\) \(y_{2}^{6}-32 y_{2}^{3}-2 y_{2}^{3} \cdot y_{1}+y_{1}^{2}+32 y_{1}+256=0\) Hence, the degree of the given equation is ' 6 '.
Jamia Millia Islamia-2013
Differential Equation
87144
The differential equation of the family of curves \(\mathrm{y}=\mathrm{ax}+\frac{1}{\mathrm{a}}\), where \(\mathrm{a}(\neq 0)\) is an arbitrary constant, has the degree
1 4
2 3
3 1
4 2
Explanation:
(D) : Given, the family of curves- \(\mathrm{y}=\mathrm{ax}+\frac{1}{\mathrm{a}} \tag{i}\) On differentiating both side, we get- \(\frac{d y}{d x}=a \cdot 1+0\) \(\mathrm{a}=\frac{\mathrm{dy}}{\mathrm{dx}}\) Put, \(a=\frac{d y}{d x}\) in equation (i), we get- \(y=\frac{d y}{d x} \cdot x+\frac{1}{\left(\frac{d y}{d x}\right)} \Rightarrow y \cdot \frac{d y}{d x}=x\left(\frac{d y}{d x}\right)^{2}+1\) \(x\left(\frac{d y}{d x}\right)^{2}-y\left(\frac{d y}{d x}\right)+1=0\) So, the differential equation of the family of curves \(y\) \(=\mathrm{ax}+\frac{1}{\mathrm{a}}\) has the degree 2 .
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Differential Equation
87132
The differential equation whose solution is \(\mathbf{A x}\) \(+B y^{2}=1\), where \(A\) and \(B\) are arbitrary constant, is of
1 first order and second degree
2 first order and first degree
3 second order and first degree
4 second order and second degree
Explanation:
(C) : Given the solution, \(a x^{2}+b y^{2}=1\) Now differentiating equation (i) w.r.t. \(x\), we get- \(2 a x+2 b y \frac{d y}{d x}=0 \text { or } \frac{y}{x}\left(\frac{d y}{d x}\right)=-\frac{a}{b}\) Again differentiating w.r.t. \(\mathrm{x}\), we get \(\frac{x\left\{\left(\frac{d y}{d x}\right)^{2}+y\left(\frac{d^{2} y}{d x^{2}}\right)-y\left(\frac{d y}{d x}\right)\right\}}{x^{2}}=0\) or, \(\quad x y\left(\frac{d^{2} y}{d x^{2}}\right)+x\left(\frac{d y}{d x}\right)^{2}-y\left(\frac{d y}{d x}\right)=0\) This is the differential equation of order 2 and degree 1.
AIEEE-2006
Differential Equation
87133
The differential equation representing the family of curves \(y^{2}=2 c(x+\sqrt{c})\), where \(c>0\), is a parameter, is of order and degree as follows
1 order 2, degree 2
2 order 1, degree 3
3 order 1, degree 1
4 order1, degree 2
Explanation:
(B) : \(\mathrm{y}^{2}=2 \mathrm{c}(\mathrm{x}+\sqrt{\mathrm{c}})\) \(y^{2}=2 c x+2 c^{\frac{3}{2}}\) On differentiating with respect to \(x\), we get- \(2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{c} \Rightarrow \mathrm{c}=\mathrm{y}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)\) On substitution equation (ii) in (i) we get- \(y^{2}=2 x y\left(\frac{d y}{d x}\right)+2\left(y \frac{d y}{d x}\right)^{\frac{3}{2}}\) \(\left\{y^{2}-2 x y\left(\frac{d y}{d x}\right)\right\}^{2}=2^{2} y^{3}\left(\frac{d y}{d x}\right)^{3}\) \(\left\{y^{2}-2 x y\left(\frac{d y}{d x}\right)\right\}^{2}=4 y^{3}\left(\frac{d y}{d x}\right)^{3}\) order \(=\) highest differential equation Degree \(=\) highest degree of order \(\therefore\) order \(=1\) and degree \(=3\)
Jamia Millia Cslamia-2012
Differential Equation
87134
The degree of the differential equation \(y_{2}^{3 / 2}-y_{1}^{1 / 2}-4=0\) is
1 6
2 3
3 2
4 4
Explanation:
(A) : \(\mathrm{y}_{2}^{3 / 2}=\mathrm{y}_{1}^{1 / 2}+4\) Squaring on both sides we get \(y_{2}^{3}=y_{1}+16+8 y_{1}^{1 / 2}\) \(\left(y_{2}^{3}-y_{1}-16\right)^{2}=64 y_{1}\) \(y_{2}^{6}-32 y_{2}^{3}-2 y_{2}^{3} \cdot y_{1}+y_{1}^{2}+32 y_{1}+256=0\) Hence, the degree of the given equation is ' 6 '.
Jamia Millia Islamia-2013
Differential Equation
87144
The differential equation of the family of curves \(\mathrm{y}=\mathrm{ax}+\frac{1}{\mathrm{a}}\), where \(\mathrm{a}(\neq 0)\) is an arbitrary constant, has the degree
1 4
2 3
3 1
4 2
Explanation:
(D) : Given, the family of curves- \(\mathrm{y}=\mathrm{ax}+\frac{1}{\mathrm{a}} \tag{i}\) On differentiating both side, we get- \(\frac{d y}{d x}=a \cdot 1+0\) \(\mathrm{a}=\frac{\mathrm{dy}}{\mathrm{dx}}\) Put, \(a=\frac{d y}{d x}\) in equation (i), we get- \(y=\frac{d y}{d x} \cdot x+\frac{1}{\left(\frac{d y}{d x}\right)} \Rightarrow y \cdot \frac{d y}{d x}=x\left(\frac{d y}{d x}\right)^{2}+1\) \(x\left(\frac{d y}{d x}\right)^{2}-y\left(\frac{d y}{d x}\right)+1=0\) So, the differential equation of the family of curves \(y\) \(=\mathrm{ax}+\frac{1}{\mathrm{a}}\) has the degree 2 .
87132
The differential equation whose solution is \(\mathbf{A x}\) \(+B y^{2}=1\), where \(A\) and \(B\) are arbitrary constant, is of
1 first order and second degree
2 first order and first degree
3 second order and first degree
4 second order and second degree
Explanation:
(C) : Given the solution, \(a x^{2}+b y^{2}=1\) Now differentiating equation (i) w.r.t. \(x\), we get- \(2 a x+2 b y \frac{d y}{d x}=0 \text { or } \frac{y}{x}\left(\frac{d y}{d x}\right)=-\frac{a}{b}\) Again differentiating w.r.t. \(\mathrm{x}\), we get \(\frac{x\left\{\left(\frac{d y}{d x}\right)^{2}+y\left(\frac{d^{2} y}{d x^{2}}\right)-y\left(\frac{d y}{d x}\right)\right\}}{x^{2}}=0\) or, \(\quad x y\left(\frac{d^{2} y}{d x^{2}}\right)+x\left(\frac{d y}{d x}\right)^{2}-y\left(\frac{d y}{d x}\right)=0\) This is the differential equation of order 2 and degree 1.
AIEEE-2006
Differential Equation
87133
The differential equation representing the family of curves \(y^{2}=2 c(x+\sqrt{c})\), where \(c>0\), is a parameter, is of order and degree as follows
1 order 2, degree 2
2 order 1, degree 3
3 order 1, degree 1
4 order1, degree 2
Explanation:
(B) : \(\mathrm{y}^{2}=2 \mathrm{c}(\mathrm{x}+\sqrt{\mathrm{c}})\) \(y^{2}=2 c x+2 c^{\frac{3}{2}}\) On differentiating with respect to \(x\), we get- \(2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{c} \Rightarrow \mathrm{c}=\mathrm{y}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)\) On substitution equation (ii) in (i) we get- \(y^{2}=2 x y\left(\frac{d y}{d x}\right)+2\left(y \frac{d y}{d x}\right)^{\frac{3}{2}}\) \(\left\{y^{2}-2 x y\left(\frac{d y}{d x}\right)\right\}^{2}=2^{2} y^{3}\left(\frac{d y}{d x}\right)^{3}\) \(\left\{y^{2}-2 x y\left(\frac{d y}{d x}\right)\right\}^{2}=4 y^{3}\left(\frac{d y}{d x}\right)^{3}\) order \(=\) highest differential equation Degree \(=\) highest degree of order \(\therefore\) order \(=1\) and degree \(=3\)
Jamia Millia Cslamia-2012
Differential Equation
87134
The degree of the differential equation \(y_{2}^{3 / 2}-y_{1}^{1 / 2}-4=0\) is
1 6
2 3
3 2
4 4
Explanation:
(A) : \(\mathrm{y}_{2}^{3 / 2}=\mathrm{y}_{1}^{1 / 2}+4\) Squaring on both sides we get \(y_{2}^{3}=y_{1}+16+8 y_{1}^{1 / 2}\) \(\left(y_{2}^{3}-y_{1}-16\right)^{2}=64 y_{1}\) \(y_{2}^{6}-32 y_{2}^{3}-2 y_{2}^{3} \cdot y_{1}+y_{1}^{2}+32 y_{1}+256=0\) Hence, the degree of the given equation is ' 6 '.
Jamia Millia Islamia-2013
Differential Equation
87144
The differential equation of the family of curves \(\mathrm{y}=\mathrm{ax}+\frac{1}{\mathrm{a}}\), where \(\mathrm{a}(\neq 0)\) is an arbitrary constant, has the degree
1 4
2 3
3 1
4 2
Explanation:
(D) : Given, the family of curves- \(\mathrm{y}=\mathrm{ax}+\frac{1}{\mathrm{a}} \tag{i}\) On differentiating both side, we get- \(\frac{d y}{d x}=a \cdot 1+0\) \(\mathrm{a}=\frac{\mathrm{dy}}{\mathrm{dx}}\) Put, \(a=\frac{d y}{d x}\) in equation (i), we get- \(y=\frac{d y}{d x} \cdot x+\frac{1}{\left(\frac{d y}{d x}\right)} \Rightarrow y \cdot \frac{d y}{d x}=x\left(\frac{d y}{d x}\right)^{2}+1\) \(x\left(\frac{d y}{d x}\right)^{2}-y\left(\frac{d y}{d x}\right)+1=0\) So, the differential equation of the family of curves \(y\) \(=\mathrm{ax}+\frac{1}{\mathrm{a}}\) has the degree 2 .
87132
The differential equation whose solution is \(\mathbf{A x}\) \(+B y^{2}=1\), where \(A\) and \(B\) are arbitrary constant, is of
1 first order and second degree
2 first order and first degree
3 second order and first degree
4 second order and second degree
Explanation:
(C) : Given the solution, \(a x^{2}+b y^{2}=1\) Now differentiating equation (i) w.r.t. \(x\), we get- \(2 a x+2 b y \frac{d y}{d x}=0 \text { or } \frac{y}{x}\left(\frac{d y}{d x}\right)=-\frac{a}{b}\) Again differentiating w.r.t. \(\mathrm{x}\), we get \(\frac{x\left\{\left(\frac{d y}{d x}\right)^{2}+y\left(\frac{d^{2} y}{d x^{2}}\right)-y\left(\frac{d y}{d x}\right)\right\}}{x^{2}}=0\) or, \(\quad x y\left(\frac{d^{2} y}{d x^{2}}\right)+x\left(\frac{d y}{d x}\right)^{2}-y\left(\frac{d y}{d x}\right)=0\) This is the differential equation of order 2 and degree 1.
AIEEE-2006
Differential Equation
87133
The differential equation representing the family of curves \(y^{2}=2 c(x+\sqrt{c})\), where \(c>0\), is a parameter, is of order and degree as follows
1 order 2, degree 2
2 order 1, degree 3
3 order 1, degree 1
4 order1, degree 2
Explanation:
(B) : \(\mathrm{y}^{2}=2 \mathrm{c}(\mathrm{x}+\sqrt{\mathrm{c}})\) \(y^{2}=2 c x+2 c^{\frac{3}{2}}\) On differentiating with respect to \(x\), we get- \(2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{c} \Rightarrow \mathrm{c}=\mathrm{y}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)\) On substitution equation (ii) in (i) we get- \(y^{2}=2 x y\left(\frac{d y}{d x}\right)+2\left(y \frac{d y}{d x}\right)^{\frac{3}{2}}\) \(\left\{y^{2}-2 x y\left(\frac{d y}{d x}\right)\right\}^{2}=2^{2} y^{3}\left(\frac{d y}{d x}\right)^{3}\) \(\left\{y^{2}-2 x y\left(\frac{d y}{d x}\right)\right\}^{2}=4 y^{3}\left(\frac{d y}{d x}\right)^{3}\) order \(=\) highest differential equation Degree \(=\) highest degree of order \(\therefore\) order \(=1\) and degree \(=3\)
Jamia Millia Cslamia-2012
Differential Equation
87134
The degree of the differential equation \(y_{2}^{3 / 2}-y_{1}^{1 / 2}-4=0\) is
1 6
2 3
3 2
4 4
Explanation:
(A) : \(\mathrm{y}_{2}^{3 / 2}=\mathrm{y}_{1}^{1 / 2}+4\) Squaring on both sides we get \(y_{2}^{3}=y_{1}+16+8 y_{1}^{1 / 2}\) \(\left(y_{2}^{3}-y_{1}-16\right)^{2}=64 y_{1}\) \(y_{2}^{6}-32 y_{2}^{3}-2 y_{2}^{3} \cdot y_{1}+y_{1}^{2}+32 y_{1}+256=0\) Hence, the degree of the given equation is ' 6 '.
Jamia Millia Islamia-2013
Differential Equation
87144
The differential equation of the family of curves \(\mathrm{y}=\mathrm{ax}+\frac{1}{\mathrm{a}}\), where \(\mathrm{a}(\neq 0)\) is an arbitrary constant, has the degree
1 4
2 3
3 1
4 2
Explanation:
(D) : Given, the family of curves- \(\mathrm{y}=\mathrm{ax}+\frac{1}{\mathrm{a}} \tag{i}\) On differentiating both side, we get- \(\frac{d y}{d x}=a \cdot 1+0\) \(\mathrm{a}=\frac{\mathrm{dy}}{\mathrm{dx}}\) Put, \(a=\frac{d y}{d x}\) in equation (i), we get- \(y=\frac{d y}{d x} \cdot x+\frac{1}{\left(\frac{d y}{d x}\right)} \Rightarrow y \cdot \frac{d y}{d x}=x\left(\frac{d y}{d x}\right)^{2}+1\) \(x\left(\frac{d y}{d x}\right)^{2}-y\left(\frac{d y}{d x}\right)+1=0\) So, the differential equation of the family of curves \(y\) \(=\mathrm{ax}+\frac{1}{\mathrm{a}}\) has the degree 2 .
