NEET Test Series from KOTA - 10 Papers In MS WORD
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Differential Equation
87589
The slope of a curve at any point is the reciprocal of twice the ordinate of the point and it passes through the point \((4,3)\). The equation of the curve is
1 \(x_{2}^{2}=y+5\)
2 \(y_{2}^{2}=x-5\)
3 \(y^{2}=x+5\)
4 \(x^{2}=y-5\)
Explanation:
(C) : Given, Slope \(=\frac{d y}{d x} \Rightarrow \frac{d y}{d x}=\frac{1}{2 y} \Rightarrow 2 y d y=d x\) Integrating both sides, we get \(\mathrm{y}^{2}=\mathrm{x}+\mathrm{c}\) This passes through \((4,3)\) \(\therefore 9=4+\mathrm{c} \Rightarrow \mathrm{c}=5\) So, the equation of curve is \(\mathrm{y}^{2}=\mathrm{x}+5\)
SRM JEEE-2011
Differential Equation
87590
An integrating factor of the differential equation \(\sin x \frac{d y}{d x}+2 y \cos x=1\) is
1 \(\sin ^{2} \mathrm{x}\)
2 \(\frac{2}{\sin \mathrm{x}}\)
3 \(\log |\sin x|\)
4 \(\frac{1}{\sin ^{2} x}\)
Explanation:
(A) : Given, Differential equation is \(\sin x \frac{d y}{d x}+2 y \cos x=1\) \(\Rightarrow \frac{d y}{d x}+2 y \frac{\cos x}{\sin x}=\frac{1}{\sin x}\) \(\Rightarrow \frac{d y}{d x}+(2 \cot x) y=\operatorname{cosec} x\) Which is form of \(\frac{d y}{d x}+P y=Q\) \(P=2 \cot x \& Q=\operatorname{cosec} x\) I.F \(=\mathrm{e}^{\int 2 \cot x \mathrm{dx}}\) \(=\mathrm{e}^{2 \log (\sin \mathrm{x})}=\sin ^{2} \mathrm{x}\)
BITSAT-2018
Differential Equation
87591
The solution of \(\frac{d y}{d x}=\cos (x+y)+\sin (x+y)\) is
87589
The slope of a curve at any point is the reciprocal of twice the ordinate of the point and it passes through the point \((4,3)\). The equation of the curve is
1 \(x_{2}^{2}=y+5\)
2 \(y_{2}^{2}=x-5\)
3 \(y^{2}=x+5\)
4 \(x^{2}=y-5\)
Explanation:
(C) : Given, Slope \(=\frac{d y}{d x} \Rightarrow \frac{d y}{d x}=\frac{1}{2 y} \Rightarrow 2 y d y=d x\) Integrating both sides, we get \(\mathrm{y}^{2}=\mathrm{x}+\mathrm{c}\) This passes through \((4,3)\) \(\therefore 9=4+\mathrm{c} \Rightarrow \mathrm{c}=5\) So, the equation of curve is \(\mathrm{y}^{2}=\mathrm{x}+5\)
SRM JEEE-2011
Differential Equation
87590
An integrating factor of the differential equation \(\sin x \frac{d y}{d x}+2 y \cos x=1\) is
1 \(\sin ^{2} \mathrm{x}\)
2 \(\frac{2}{\sin \mathrm{x}}\)
3 \(\log |\sin x|\)
4 \(\frac{1}{\sin ^{2} x}\)
Explanation:
(A) : Given, Differential equation is \(\sin x \frac{d y}{d x}+2 y \cos x=1\) \(\Rightarrow \frac{d y}{d x}+2 y \frac{\cos x}{\sin x}=\frac{1}{\sin x}\) \(\Rightarrow \frac{d y}{d x}+(2 \cot x) y=\operatorname{cosec} x\) Which is form of \(\frac{d y}{d x}+P y=Q\) \(P=2 \cot x \& Q=\operatorname{cosec} x\) I.F \(=\mathrm{e}^{\int 2 \cot x \mathrm{dx}}\) \(=\mathrm{e}^{2 \log (\sin \mathrm{x})}=\sin ^{2} \mathrm{x}\)
BITSAT-2018
Differential Equation
87591
The solution of \(\frac{d y}{d x}=\cos (x+y)+\sin (x+y)\) is
87589
The slope of a curve at any point is the reciprocal of twice the ordinate of the point and it passes through the point \((4,3)\). The equation of the curve is
1 \(x_{2}^{2}=y+5\)
2 \(y_{2}^{2}=x-5\)
3 \(y^{2}=x+5\)
4 \(x^{2}=y-5\)
Explanation:
(C) : Given, Slope \(=\frac{d y}{d x} \Rightarrow \frac{d y}{d x}=\frac{1}{2 y} \Rightarrow 2 y d y=d x\) Integrating both sides, we get \(\mathrm{y}^{2}=\mathrm{x}+\mathrm{c}\) This passes through \((4,3)\) \(\therefore 9=4+\mathrm{c} \Rightarrow \mathrm{c}=5\) So, the equation of curve is \(\mathrm{y}^{2}=\mathrm{x}+5\)
SRM JEEE-2011
Differential Equation
87590
An integrating factor of the differential equation \(\sin x \frac{d y}{d x}+2 y \cos x=1\) is
1 \(\sin ^{2} \mathrm{x}\)
2 \(\frac{2}{\sin \mathrm{x}}\)
3 \(\log |\sin x|\)
4 \(\frac{1}{\sin ^{2} x}\)
Explanation:
(A) : Given, Differential equation is \(\sin x \frac{d y}{d x}+2 y \cos x=1\) \(\Rightarrow \frac{d y}{d x}+2 y \frac{\cos x}{\sin x}=\frac{1}{\sin x}\) \(\Rightarrow \frac{d y}{d x}+(2 \cot x) y=\operatorname{cosec} x\) Which is form of \(\frac{d y}{d x}+P y=Q\) \(P=2 \cot x \& Q=\operatorname{cosec} x\) I.F \(=\mathrm{e}^{\int 2 \cot x \mathrm{dx}}\) \(=\mathrm{e}^{2 \log (\sin \mathrm{x})}=\sin ^{2} \mathrm{x}\)
BITSAT-2018
Differential Equation
87591
The solution of \(\frac{d y}{d x}=\cos (x+y)+\sin (x+y)\) is
87589
The slope of a curve at any point is the reciprocal of twice the ordinate of the point and it passes through the point \((4,3)\). The equation of the curve is
1 \(x_{2}^{2}=y+5\)
2 \(y_{2}^{2}=x-5\)
3 \(y^{2}=x+5\)
4 \(x^{2}=y-5\)
Explanation:
(C) : Given, Slope \(=\frac{d y}{d x} \Rightarrow \frac{d y}{d x}=\frac{1}{2 y} \Rightarrow 2 y d y=d x\) Integrating both sides, we get \(\mathrm{y}^{2}=\mathrm{x}+\mathrm{c}\) This passes through \((4,3)\) \(\therefore 9=4+\mathrm{c} \Rightarrow \mathrm{c}=5\) So, the equation of curve is \(\mathrm{y}^{2}=\mathrm{x}+5\)
SRM JEEE-2011
Differential Equation
87590
An integrating factor of the differential equation \(\sin x \frac{d y}{d x}+2 y \cos x=1\) is
1 \(\sin ^{2} \mathrm{x}\)
2 \(\frac{2}{\sin \mathrm{x}}\)
3 \(\log |\sin x|\)
4 \(\frac{1}{\sin ^{2} x}\)
Explanation:
(A) : Given, Differential equation is \(\sin x \frac{d y}{d x}+2 y \cos x=1\) \(\Rightarrow \frac{d y}{d x}+2 y \frac{\cos x}{\sin x}=\frac{1}{\sin x}\) \(\Rightarrow \frac{d y}{d x}+(2 \cot x) y=\operatorname{cosec} x\) Which is form of \(\frac{d y}{d x}+P y=Q\) \(P=2 \cot x \& Q=\operatorname{cosec} x\) I.F \(=\mathrm{e}^{\int 2 \cot x \mathrm{dx}}\) \(=\mathrm{e}^{2 \log (\sin \mathrm{x})}=\sin ^{2} \mathrm{x}\)
BITSAT-2018
Differential Equation
87591
The solution of \(\frac{d y}{d x}=\cos (x+y)+\sin (x+y)\) is
