87249
The solution of the given differential equation \(\frac{d y}{d x}+2 x y=y\) is :
1 \(y=c e^{x-x^{2}}\)
2 \(y=c e^{x^{2}-x}\)
3 \(y=c e^{x}\)
4 \(y=c e^{-x^{2}}\)
Explanation:
(A) : Given, Differential equation, \(\frac{d y}{d x}+2 x y=y \Rightarrow \frac{d y}{d x}+(2 x-1) y=0\) \(\frac{d y}{y}=(1-2 x) d x\) Integrating on both sides, we get- \(\int \frac{d y}{y}=\int(1-2 x) d x \Rightarrow \log y=x-2 \frac{x^{2}}{2}+c\) \(\log y+\left(x^{2}-x\right)=\log c \Rightarrow \log \frac{c}{y}=x^{2}-x\) \(\mathrm{c}=\mathrm{ye}^{\mathrm{x}^{2}-\mathrm{x}}, \quad \mathrm{y}=\mathrm{ce}^{\mathrm{x}-\mathrm{x}^{2}}\)
BCECE-2003
Differential Equation
87250
The differential equation of all circles through the origin and having their centers on the \(x\) axis is
1 \(x^{2}-y^{2}=2 \frac{d y}{d x}\)
2 \(x^{2}-y^{2}=2 x y \frac{d y}{d x}\)
3 \(y^{2}-x^{2}=2 x y \frac{d y}{d x}\)
4 \(y^{2}-x^{2}=2 \frac{d y}{d x}\)
Explanation:
(C) : General equation of circle is, \(x^{2}+y^{2}+2 g x+2 f y+c=0\) Center lies on \(\mathrm{x}\) axis and passes through origin. Then centre of circle \((-\mathrm{g}, 0)\). Equation of circle passing through origin and centre on \(x\) axis is \(x^{2}+y^{2}+2 g x=0 \tag{i}\\)\(g=-\frac{x^{2}+y^{2}}{2 x}\) Differentiating w.r.t \(x\) in (i), we get- \(2 x+2 y \frac{d y}{d x}+2 g=0\) Putting the value of \(g\), we get- \(2 x+2 y \frac{d y}{d x}+2\left[-\frac{\left(x^{2}+y^{2}\right)}{2 x}\right]=0\) \(2 x+2 y \frac{d y}{d x}-\frac{\left(x^{2}+y^{2}\right)}{x}=0\) Multiplied by ' \(x\) ' both sides, \(2 x^{2}+2 x y \frac{d y}{d x}=x^{2}+y^{2} \Rightarrow x^{2}+2 x y \frac{d y}{d x}=y^{2}\) \(y^{2}-x^{2}=2 x y \frac{d y}{d x}\)
AMU-2006
Differential Equation
87251
The differential equation \(y^{\prime \prime}+k y^{\prime}+4 y=0\) has solution of the form \(y=A e^{a x} \cos b x+B e^{a x} \sin b x\) for all values of \(k\), if
1 \(-4\lt \mathrm{k}\lt 4\)
2 \(\mathrm{k}\lt -4, \mathrm{k}>4\)
3 \(\mathrm{k}=0\) or 4
4 None of the above
Explanation:
(A) : The given differentiating equation is, \(\mathrm{y}^{\prime \prime}+\mathrm{ky} \mathrm{y}^{\prime}+4 \mathrm{y}=0\) \(\frac{d^{2} y}{d x^{2}}+k \frac{d y}{d x}+4 y=0\) We that, \(\frac{\mathrm{d}^{2}}{\mathrm{dx}^{2}}=\mathrm{D}^{2} \& \frac{\mathrm{d}}{\mathrm{dx}}=\mathrm{D}\) So, we get:- \(\mathrm{D}^{2} \mathrm{y}+\mathrm{kDy}+4 \mathrm{y}=0 \Rightarrow \mathrm{D}^{2}+\mathrm{kD}+4=0\) So, characteristic equation will be \(b^{2}-4 a c \Rightarrow \quad=k^{2}-4 .(4) \tag{i}\) Now, solution is given as:- \(\mathrm{y}=\mathrm{Ae}^{\mathrm{ax}} \cos \mathrm{bx}+\mathrm{Be}^{\mathrm{ax}} \sin \mathrm{bx}\) \(=\mathrm{e}^{\mathrm{ax}}(\mathrm{A} \cos b \mathrm{x}+\mathrm{B} \sin \mathrm{bx})\) Since this is the standard solution of complex conjugate root i.e. a \(\pm\) bi And this roots come when characteristic equation is \(\lt 0\) i.e. \(\mathrm{k}^{2}-16\lt 0\) or \(\mathrm{k}^{2}\lt 16\) or \(-4\lt \mathrm{k}\lt 4\) So, range of \(\mathrm{k}\) is \((-4,4)\)
SCRA-2009
Differential Equation
87252
The solution of the differential equation \(\frac{d y}{d x}-\frac{y}{x}=1 \text { is }\)
1 \(x^{2} \log _{e} x+y=c\)
2 \(x \log _{e} x+c x=y\)
3 \(x^{2} \log _{e} x-y=c\)
4 \(x \log _{e} x+y=c x\)
Explanation:
(B) : Given, Differential equation, \(\frac{\mathrm{dy}}{\mathrm{dx}}-\frac{\mathrm{y}}{\mathrm{x}}=1\) Which is the form of \(\frac{d y}{d x}+P y=Q\) \(P=-\frac{1}{x} \text { and } Q=1\) \(I \cdot F=e^{\int P d x}=e^{\int \frac{-1}{x} d x}=e^{-\log x}=\frac{1}{x}\) For general solution, we get - \(\mathrm{y} \cdot \mathrm{I} \cdot \mathrm{F}=\int \mathrm{I} \cdot \mathrm{F} \times \mathrm{Qdx}+\mathrm{c} \Rightarrow \mathrm{y} \cdot \frac{1}{\mathrm{x}}=\int \frac{1}{\mathrm{x}} \cdot \mathrm{dx}+\mathrm{c}\) \(\frac{y}{x}=\log x+c \Rightarrow y=x \log x+x c\) \(x \log _{e} x+c x=y\)
87249
The solution of the given differential equation \(\frac{d y}{d x}+2 x y=y\) is :
1 \(y=c e^{x-x^{2}}\)
2 \(y=c e^{x^{2}-x}\)
3 \(y=c e^{x}\)
4 \(y=c e^{-x^{2}}\)
Explanation:
(A) : Given, Differential equation, \(\frac{d y}{d x}+2 x y=y \Rightarrow \frac{d y}{d x}+(2 x-1) y=0\) \(\frac{d y}{y}=(1-2 x) d x\) Integrating on both sides, we get- \(\int \frac{d y}{y}=\int(1-2 x) d x \Rightarrow \log y=x-2 \frac{x^{2}}{2}+c\) \(\log y+\left(x^{2}-x\right)=\log c \Rightarrow \log \frac{c}{y}=x^{2}-x\) \(\mathrm{c}=\mathrm{ye}^{\mathrm{x}^{2}-\mathrm{x}}, \quad \mathrm{y}=\mathrm{ce}^{\mathrm{x}-\mathrm{x}^{2}}\)
BCECE-2003
Differential Equation
87250
The differential equation of all circles through the origin and having their centers on the \(x\) axis is
1 \(x^{2}-y^{2}=2 \frac{d y}{d x}\)
2 \(x^{2}-y^{2}=2 x y \frac{d y}{d x}\)
3 \(y^{2}-x^{2}=2 x y \frac{d y}{d x}\)
4 \(y^{2}-x^{2}=2 \frac{d y}{d x}\)
Explanation:
(C) : General equation of circle is, \(x^{2}+y^{2}+2 g x+2 f y+c=0\) Center lies on \(\mathrm{x}\) axis and passes through origin. Then centre of circle \((-\mathrm{g}, 0)\). Equation of circle passing through origin and centre on \(x\) axis is \(x^{2}+y^{2}+2 g x=0 \tag{i}\\)\(g=-\frac{x^{2}+y^{2}}{2 x}\) Differentiating w.r.t \(x\) in (i), we get- \(2 x+2 y \frac{d y}{d x}+2 g=0\) Putting the value of \(g\), we get- \(2 x+2 y \frac{d y}{d x}+2\left[-\frac{\left(x^{2}+y^{2}\right)}{2 x}\right]=0\) \(2 x+2 y \frac{d y}{d x}-\frac{\left(x^{2}+y^{2}\right)}{x}=0\) Multiplied by ' \(x\) ' both sides, \(2 x^{2}+2 x y \frac{d y}{d x}=x^{2}+y^{2} \Rightarrow x^{2}+2 x y \frac{d y}{d x}=y^{2}\) \(y^{2}-x^{2}=2 x y \frac{d y}{d x}\)
AMU-2006
Differential Equation
87251
The differential equation \(y^{\prime \prime}+k y^{\prime}+4 y=0\) has solution of the form \(y=A e^{a x} \cos b x+B e^{a x} \sin b x\) for all values of \(k\), if
1 \(-4\lt \mathrm{k}\lt 4\)
2 \(\mathrm{k}\lt -4, \mathrm{k}>4\)
3 \(\mathrm{k}=0\) or 4
4 None of the above
Explanation:
(A) : The given differentiating equation is, \(\mathrm{y}^{\prime \prime}+\mathrm{ky} \mathrm{y}^{\prime}+4 \mathrm{y}=0\) \(\frac{d^{2} y}{d x^{2}}+k \frac{d y}{d x}+4 y=0\) We that, \(\frac{\mathrm{d}^{2}}{\mathrm{dx}^{2}}=\mathrm{D}^{2} \& \frac{\mathrm{d}}{\mathrm{dx}}=\mathrm{D}\) So, we get:- \(\mathrm{D}^{2} \mathrm{y}+\mathrm{kDy}+4 \mathrm{y}=0 \Rightarrow \mathrm{D}^{2}+\mathrm{kD}+4=0\) So, characteristic equation will be \(b^{2}-4 a c \Rightarrow \quad=k^{2}-4 .(4) \tag{i}\) Now, solution is given as:- \(\mathrm{y}=\mathrm{Ae}^{\mathrm{ax}} \cos \mathrm{bx}+\mathrm{Be}^{\mathrm{ax}} \sin \mathrm{bx}\) \(=\mathrm{e}^{\mathrm{ax}}(\mathrm{A} \cos b \mathrm{x}+\mathrm{B} \sin \mathrm{bx})\) Since this is the standard solution of complex conjugate root i.e. a \(\pm\) bi And this roots come when characteristic equation is \(\lt 0\) i.e. \(\mathrm{k}^{2}-16\lt 0\) or \(\mathrm{k}^{2}\lt 16\) or \(-4\lt \mathrm{k}\lt 4\) So, range of \(\mathrm{k}\) is \((-4,4)\)
SCRA-2009
Differential Equation
87252
The solution of the differential equation \(\frac{d y}{d x}-\frac{y}{x}=1 \text { is }\)
1 \(x^{2} \log _{e} x+y=c\)
2 \(x \log _{e} x+c x=y\)
3 \(x^{2} \log _{e} x-y=c\)
4 \(x \log _{e} x+y=c x\)
Explanation:
(B) : Given, Differential equation, \(\frac{\mathrm{dy}}{\mathrm{dx}}-\frac{\mathrm{y}}{\mathrm{x}}=1\) Which is the form of \(\frac{d y}{d x}+P y=Q\) \(P=-\frac{1}{x} \text { and } Q=1\) \(I \cdot F=e^{\int P d x}=e^{\int \frac{-1}{x} d x}=e^{-\log x}=\frac{1}{x}\) For general solution, we get - \(\mathrm{y} \cdot \mathrm{I} \cdot \mathrm{F}=\int \mathrm{I} \cdot \mathrm{F} \times \mathrm{Qdx}+\mathrm{c} \Rightarrow \mathrm{y} \cdot \frac{1}{\mathrm{x}}=\int \frac{1}{\mathrm{x}} \cdot \mathrm{dx}+\mathrm{c}\) \(\frac{y}{x}=\log x+c \Rightarrow y=x \log x+x c\) \(x \log _{e} x+c x=y\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Differential Equation
87249
The solution of the given differential equation \(\frac{d y}{d x}+2 x y=y\) is :
1 \(y=c e^{x-x^{2}}\)
2 \(y=c e^{x^{2}-x}\)
3 \(y=c e^{x}\)
4 \(y=c e^{-x^{2}}\)
Explanation:
(A) : Given, Differential equation, \(\frac{d y}{d x}+2 x y=y \Rightarrow \frac{d y}{d x}+(2 x-1) y=0\) \(\frac{d y}{y}=(1-2 x) d x\) Integrating on both sides, we get- \(\int \frac{d y}{y}=\int(1-2 x) d x \Rightarrow \log y=x-2 \frac{x^{2}}{2}+c\) \(\log y+\left(x^{2}-x\right)=\log c \Rightarrow \log \frac{c}{y}=x^{2}-x\) \(\mathrm{c}=\mathrm{ye}^{\mathrm{x}^{2}-\mathrm{x}}, \quad \mathrm{y}=\mathrm{ce}^{\mathrm{x}-\mathrm{x}^{2}}\)
BCECE-2003
Differential Equation
87250
The differential equation of all circles through the origin and having their centers on the \(x\) axis is
1 \(x^{2}-y^{2}=2 \frac{d y}{d x}\)
2 \(x^{2}-y^{2}=2 x y \frac{d y}{d x}\)
3 \(y^{2}-x^{2}=2 x y \frac{d y}{d x}\)
4 \(y^{2}-x^{2}=2 \frac{d y}{d x}\)
Explanation:
(C) : General equation of circle is, \(x^{2}+y^{2}+2 g x+2 f y+c=0\) Center lies on \(\mathrm{x}\) axis and passes through origin. Then centre of circle \((-\mathrm{g}, 0)\). Equation of circle passing through origin and centre on \(x\) axis is \(x^{2}+y^{2}+2 g x=0 \tag{i}\\)\(g=-\frac{x^{2}+y^{2}}{2 x}\) Differentiating w.r.t \(x\) in (i), we get- \(2 x+2 y \frac{d y}{d x}+2 g=0\) Putting the value of \(g\), we get- \(2 x+2 y \frac{d y}{d x}+2\left[-\frac{\left(x^{2}+y^{2}\right)}{2 x}\right]=0\) \(2 x+2 y \frac{d y}{d x}-\frac{\left(x^{2}+y^{2}\right)}{x}=0\) Multiplied by ' \(x\) ' both sides, \(2 x^{2}+2 x y \frac{d y}{d x}=x^{2}+y^{2} \Rightarrow x^{2}+2 x y \frac{d y}{d x}=y^{2}\) \(y^{2}-x^{2}=2 x y \frac{d y}{d x}\)
AMU-2006
Differential Equation
87251
The differential equation \(y^{\prime \prime}+k y^{\prime}+4 y=0\) has solution of the form \(y=A e^{a x} \cos b x+B e^{a x} \sin b x\) for all values of \(k\), if
1 \(-4\lt \mathrm{k}\lt 4\)
2 \(\mathrm{k}\lt -4, \mathrm{k}>4\)
3 \(\mathrm{k}=0\) or 4
4 None of the above
Explanation:
(A) : The given differentiating equation is, \(\mathrm{y}^{\prime \prime}+\mathrm{ky} \mathrm{y}^{\prime}+4 \mathrm{y}=0\) \(\frac{d^{2} y}{d x^{2}}+k \frac{d y}{d x}+4 y=0\) We that, \(\frac{\mathrm{d}^{2}}{\mathrm{dx}^{2}}=\mathrm{D}^{2} \& \frac{\mathrm{d}}{\mathrm{dx}}=\mathrm{D}\) So, we get:- \(\mathrm{D}^{2} \mathrm{y}+\mathrm{kDy}+4 \mathrm{y}=0 \Rightarrow \mathrm{D}^{2}+\mathrm{kD}+4=0\) So, characteristic equation will be \(b^{2}-4 a c \Rightarrow \quad=k^{2}-4 .(4) \tag{i}\) Now, solution is given as:- \(\mathrm{y}=\mathrm{Ae}^{\mathrm{ax}} \cos \mathrm{bx}+\mathrm{Be}^{\mathrm{ax}} \sin \mathrm{bx}\) \(=\mathrm{e}^{\mathrm{ax}}(\mathrm{A} \cos b \mathrm{x}+\mathrm{B} \sin \mathrm{bx})\) Since this is the standard solution of complex conjugate root i.e. a \(\pm\) bi And this roots come when characteristic equation is \(\lt 0\) i.e. \(\mathrm{k}^{2}-16\lt 0\) or \(\mathrm{k}^{2}\lt 16\) or \(-4\lt \mathrm{k}\lt 4\) So, range of \(\mathrm{k}\) is \((-4,4)\)
SCRA-2009
Differential Equation
87252
The solution of the differential equation \(\frac{d y}{d x}-\frac{y}{x}=1 \text { is }\)
1 \(x^{2} \log _{e} x+y=c\)
2 \(x \log _{e} x+c x=y\)
3 \(x^{2} \log _{e} x-y=c\)
4 \(x \log _{e} x+y=c x\)
Explanation:
(B) : Given, Differential equation, \(\frac{\mathrm{dy}}{\mathrm{dx}}-\frac{\mathrm{y}}{\mathrm{x}}=1\) Which is the form of \(\frac{d y}{d x}+P y=Q\) \(P=-\frac{1}{x} \text { and } Q=1\) \(I \cdot F=e^{\int P d x}=e^{\int \frac{-1}{x} d x}=e^{-\log x}=\frac{1}{x}\) For general solution, we get - \(\mathrm{y} \cdot \mathrm{I} \cdot \mathrm{F}=\int \mathrm{I} \cdot \mathrm{F} \times \mathrm{Qdx}+\mathrm{c} \Rightarrow \mathrm{y} \cdot \frac{1}{\mathrm{x}}=\int \frac{1}{\mathrm{x}} \cdot \mathrm{dx}+\mathrm{c}\) \(\frac{y}{x}=\log x+c \Rightarrow y=x \log x+x c\) \(x \log _{e} x+c x=y\)
87249
The solution of the given differential equation \(\frac{d y}{d x}+2 x y=y\) is :
1 \(y=c e^{x-x^{2}}\)
2 \(y=c e^{x^{2}-x}\)
3 \(y=c e^{x}\)
4 \(y=c e^{-x^{2}}\)
Explanation:
(A) : Given, Differential equation, \(\frac{d y}{d x}+2 x y=y \Rightarrow \frac{d y}{d x}+(2 x-1) y=0\) \(\frac{d y}{y}=(1-2 x) d x\) Integrating on both sides, we get- \(\int \frac{d y}{y}=\int(1-2 x) d x \Rightarrow \log y=x-2 \frac{x^{2}}{2}+c\) \(\log y+\left(x^{2}-x\right)=\log c \Rightarrow \log \frac{c}{y}=x^{2}-x\) \(\mathrm{c}=\mathrm{ye}^{\mathrm{x}^{2}-\mathrm{x}}, \quad \mathrm{y}=\mathrm{ce}^{\mathrm{x}-\mathrm{x}^{2}}\)
BCECE-2003
Differential Equation
87250
The differential equation of all circles through the origin and having their centers on the \(x\) axis is
1 \(x^{2}-y^{2}=2 \frac{d y}{d x}\)
2 \(x^{2}-y^{2}=2 x y \frac{d y}{d x}\)
3 \(y^{2}-x^{2}=2 x y \frac{d y}{d x}\)
4 \(y^{2}-x^{2}=2 \frac{d y}{d x}\)
Explanation:
(C) : General equation of circle is, \(x^{2}+y^{2}+2 g x+2 f y+c=0\) Center lies on \(\mathrm{x}\) axis and passes through origin. Then centre of circle \((-\mathrm{g}, 0)\). Equation of circle passing through origin and centre on \(x\) axis is \(x^{2}+y^{2}+2 g x=0 \tag{i}\\)\(g=-\frac{x^{2}+y^{2}}{2 x}\) Differentiating w.r.t \(x\) in (i), we get- \(2 x+2 y \frac{d y}{d x}+2 g=0\) Putting the value of \(g\), we get- \(2 x+2 y \frac{d y}{d x}+2\left[-\frac{\left(x^{2}+y^{2}\right)}{2 x}\right]=0\) \(2 x+2 y \frac{d y}{d x}-\frac{\left(x^{2}+y^{2}\right)}{x}=0\) Multiplied by ' \(x\) ' both sides, \(2 x^{2}+2 x y \frac{d y}{d x}=x^{2}+y^{2} \Rightarrow x^{2}+2 x y \frac{d y}{d x}=y^{2}\) \(y^{2}-x^{2}=2 x y \frac{d y}{d x}\)
AMU-2006
Differential Equation
87251
The differential equation \(y^{\prime \prime}+k y^{\prime}+4 y=0\) has solution of the form \(y=A e^{a x} \cos b x+B e^{a x} \sin b x\) for all values of \(k\), if
1 \(-4\lt \mathrm{k}\lt 4\)
2 \(\mathrm{k}\lt -4, \mathrm{k}>4\)
3 \(\mathrm{k}=0\) or 4
4 None of the above
Explanation:
(A) : The given differentiating equation is, \(\mathrm{y}^{\prime \prime}+\mathrm{ky} \mathrm{y}^{\prime}+4 \mathrm{y}=0\) \(\frac{d^{2} y}{d x^{2}}+k \frac{d y}{d x}+4 y=0\) We that, \(\frac{\mathrm{d}^{2}}{\mathrm{dx}^{2}}=\mathrm{D}^{2} \& \frac{\mathrm{d}}{\mathrm{dx}}=\mathrm{D}\) So, we get:- \(\mathrm{D}^{2} \mathrm{y}+\mathrm{kDy}+4 \mathrm{y}=0 \Rightarrow \mathrm{D}^{2}+\mathrm{kD}+4=0\) So, characteristic equation will be \(b^{2}-4 a c \Rightarrow \quad=k^{2}-4 .(4) \tag{i}\) Now, solution is given as:- \(\mathrm{y}=\mathrm{Ae}^{\mathrm{ax}} \cos \mathrm{bx}+\mathrm{Be}^{\mathrm{ax}} \sin \mathrm{bx}\) \(=\mathrm{e}^{\mathrm{ax}}(\mathrm{A} \cos b \mathrm{x}+\mathrm{B} \sin \mathrm{bx})\) Since this is the standard solution of complex conjugate root i.e. a \(\pm\) bi And this roots come when characteristic equation is \(\lt 0\) i.e. \(\mathrm{k}^{2}-16\lt 0\) or \(\mathrm{k}^{2}\lt 16\) or \(-4\lt \mathrm{k}\lt 4\) So, range of \(\mathrm{k}\) is \((-4,4)\)
SCRA-2009
Differential Equation
87252
The solution of the differential equation \(\frac{d y}{d x}-\frac{y}{x}=1 \text { is }\)
1 \(x^{2} \log _{e} x+y=c\)
2 \(x \log _{e} x+c x=y\)
3 \(x^{2} \log _{e} x-y=c\)
4 \(x \log _{e} x+y=c x\)
Explanation:
(B) : Given, Differential equation, \(\frac{\mathrm{dy}}{\mathrm{dx}}-\frac{\mathrm{y}}{\mathrm{x}}=1\) Which is the form of \(\frac{d y}{d x}+P y=Q\) \(P=-\frac{1}{x} \text { and } Q=1\) \(I \cdot F=e^{\int P d x}=e^{\int \frac{-1}{x} d x}=e^{-\log x}=\frac{1}{x}\) For general solution, we get - \(\mathrm{y} \cdot \mathrm{I} \cdot \mathrm{F}=\int \mathrm{I} \cdot \mathrm{F} \times \mathrm{Qdx}+\mathrm{c} \Rightarrow \mathrm{y} \cdot \frac{1}{\mathrm{x}}=\int \frac{1}{\mathrm{x}} \cdot \mathrm{dx}+\mathrm{c}\) \(\frac{y}{x}=\log x+c \Rightarrow y=x \log x+x c\) \(x \log _{e} x+c x=y\)