87489
The differential \(y \frac{d y}{d x}+x=\) c represents
1 a family of parabolas
2 a family of circles whose centres are on the \(x\) axis
3 a family of hyperbolas
4 a family of circles whose centres are on the yaxis
Explanation:
(B) : Given, \[ \begin{array}{l} y \frac{d y}{d x}+x=c \\ y \frac{d y}{d x}=c-x \\ y \frac{d y}{d y}=(c-x) d x \end{array} \] On integration both sides, we get \[ \begin{array}{l} \int \mathrm{y} d \mathrm{~d}=\int(\mathrm{c}-\mathrm{x}) \mathrm{dx} \\ \frac{\mathrm{y}^{2}}{2}=\mathrm{cx}-\frac{\mathrm{x}^{2}}{2}+\mathrm{k} \\ \frac{\mathrm{x}^{2}}{2}+\frac{\mathrm{y}^{2}}{2}=\mathrm{cx}+\mathrm{k} \\ \mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{cx}-2 \mathrm{k}=0 \end{array} \] This equation is represent the circle whose centre are on the x -axis.
Karnataka CET-2008
Differential Equation
87490
The differential equation of the family of circles passing through the origin and having their centres on the \( x \)-axis is
1 \( x^{2}=y^{2}+x y \frac{d y}{d x} \)
2 \( x^{2}=y^{2}+3 x y \frac{d y}{d x} \)
3 \( y^{2}=x^{2}+2 x y \frac{d y}{d x} \)
4 \( y^{2}=x^{2}-2 x y \frac{d y}{d x} \)
Explanation:
(C) : Given that family circle passing through the origin and centre on \(\mathrm{x}\) - axis. Let centre be \((h, 0)\) and radius be \(h\). Now equation of family of circle, \((\mathrm{x}-\mathrm{h})^{2}+(\mathrm{y}-\mathrm{k})^{2}=\mathrm{h}^{2}\) \((x-h)^{2}+(y-0)^{2}=h^{2}\) \((\mathrm{x}-\mathrm{h})^{2}+\mathrm{y}^{2}=\mathrm{h}^{2}\) \(\mathrm{x}^{2}+\mathrm{h}^{2}-2 \mathrm{xh}+\mathrm{y}^{2}=\mathrm{h}^{2}\) \(\mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{xh}=0\) On differentiating both sides w.r.t.x, we get- \(2 x+2 y \frac{d y}{d x}-2 h=0\) \(x+y \frac{d y}{d x}=h\) On putting the value of \(h\) in equation (i) \(x^{2}+y^{2}-2 x\left(x+y \frac{d y}{d x}\right)=0\) \(\mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{x}^{2}-2 \mathrm{xy} \frac{\mathrm{dy}}{\mathrm{dx}}=0\) \(-x^{2}+y^{2}-2 x y \frac{d y}{d x}=0\) \(\mathrm{x}^{2}-\mathrm{y}^{2}+2 \mathrm{xy} \frac{\mathrm{dy}}{\mathrm{dx}}=0\) \(y^{2}=x^{2}+2 x y \frac{d y}{d x}\)
Karnataka CET-2009
Differential Equation
87491
The integrating factor of the differential equation \(\sin y\left(\frac{d y}{d x}\right)=\cos y(1-x \cos y)\) is
1 \(\mathrm{e}^{-\mathrm{x}}\)
2 \(\mathrm{e}^{-\mathrm{y}}\)
3 \(e^{\sin y}\)
4 \(e^{\cos y}\)
Explanation:
(A) : Given, \(\sin y\left(\frac{d y}{d x}\right)=\cos y(1-x \cos y)\) \(\frac{\sin y}{\cos y}\left(\frac{d y}{d x}\right)=1-x \cos y\) \(\tan y\left(\frac{d y}{d x}\right)=1-x \cos y\) \(\tan y\left(\frac{d y}{d x}\right)=1-\frac{x}{\sec y}\) \(\tan y \cdot \sec y\left(\frac{d y}{d x}\right)=\sec y-x\) Put, \(\quad \sec \mathrm{y}=\mathrm{t}\) \(\sec y \cdot \tan y\left(\frac{d y}{d x}\right)=\frac{d t}{d x}\) Then equation becomes, \(\frac{\mathrm{dt}}{\mathrm{dx}}-\mathrm{t}=-\mathrm{x}\) Which is the form of \(\frac{d y}{d x}+P y=Q\) \(\mathrm{P}=-1 \quad\) and \(\quad \mathrm{Q}=-\mathrm{X}\) I.F. \(=\mathrm{e}^{\int-1 \mathrm{dx}}=\mathrm{e}^{-\mathrm{x}}\)
MHT CET-2020
Differential Equation
87493
Solution of the differential equation \(\frac{d y}{d x}+2 y=e^{-x}\) is
87494
The integrating factor of the differential equation \( \frac{d y}{d x}+\frac{1}{x} y=x^{3}-3 \) is
1 -y
2 x
3 -x
4 y
Explanation:
(B) : Given, \(\frac{d y}{d x}+\frac{1}{x} y=x^{3}-3\) The above differential equation of the form \(\frac{d y}{d x}+P y=Q\) \(\mathrm{P}=\frac{1}{\mathrm{x}}\) and \(\mathrm{Q}=\mathrm{x}^{3}-3\) I. F. \(=\mathrm{e}^{\int P \mathrm{Pdx}}\) \(=\mathrm{e}^{\int \frac{1}{x} \mathrm{dx}}\) \(=\mathrm{e}^{\log \mathrm{x}}=\mathrm{x}\)
87489
The differential \(y \frac{d y}{d x}+x=\) c represents
1 a family of parabolas
2 a family of circles whose centres are on the \(x\) axis
3 a family of hyperbolas
4 a family of circles whose centres are on the yaxis
Explanation:
(B) : Given, \[ \begin{array}{l} y \frac{d y}{d x}+x=c \\ y \frac{d y}{d x}=c-x \\ y \frac{d y}{d y}=(c-x) d x \end{array} \] On integration both sides, we get \[ \begin{array}{l} \int \mathrm{y} d \mathrm{~d}=\int(\mathrm{c}-\mathrm{x}) \mathrm{dx} \\ \frac{\mathrm{y}^{2}}{2}=\mathrm{cx}-\frac{\mathrm{x}^{2}}{2}+\mathrm{k} \\ \frac{\mathrm{x}^{2}}{2}+\frac{\mathrm{y}^{2}}{2}=\mathrm{cx}+\mathrm{k} \\ \mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{cx}-2 \mathrm{k}=0 \end{array} \] This equation is represent the circle whose centre are on the x -axis.
Karnataka CET-2008
Differential Equation
87490
The differential equation of the family of circles passing through the origin and having their centres on the \( x \)-axis is
1 \( x^{2}=y^{2}+x y \frac{d y}{d x} \)
2 \( x^{2}=y^{2}+3 x y \frac{d y}{d x} \)
3 \( y^{2}=x^{2}+2 x y \frac{d y}{d x} \)
4 \( y^{2}=x^{2}-2 x y \frac{d y}{d x} \)
Explanation:
(C) : Given that family circle passing through the origin and centre on \(\mathrm{x}\) - axis. Let centre be \((h, 0)\) and radius be \(h\). Now equation of family of circle, \((\mathrm{x}-\mathrm{h})^{2}+(\mathrm{y}-\mathrm{k})^{2}=\mathrm{h}^{2}\) \((x-h)^{2}+(y-0)^{2}=h^{2}\) \((\mathrm{x}-\mathrm{h})^{2}+\mathrm{y}^{2}=\mathrm{h}^{2}\) \(\mathrm{x}^{2}+\mathrm{h}^{2}-2 \mathrm{xh}+\mathrm{y}^{2}=\mathrm{h}^{2}\) \(\mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{xh}=0\) On differentiating both sides w.r.t.x, we get- \(2 x+2 y \frac{d y}{d x}-2 h=0\) \(x+y \frac{d y}{d x}=h\) On putting the value of \(h\) in equation (i) \(x^{2}+y^{2}-2 x\left(x+y \frac{d y}{d x}\right)=0\) \(\mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{x}^{2}-2 \mathrm{xy} \frac{\mathrm{dy}}{\mathrm{dx}}=0\) \(-x^{2}+y^{2}-2 x y \frac{d y}{d x}=0\) \(\mathrm{x}^{2}-\mathrm{y}^{2}+2 \mathrm{xy} \frac{\mathrm{dy}}{\mathrm{dx}}=0\) \(y^{2}=x^{2}+2 x y \frac{d y}{d x}\)
Karnataka CET-2009
Differential Equation
87491
The integrating factor of the differential equation \(\sin y\left(\frac{d y}{d x}\right)=\cos y(1-x \cos y)\) is
1 \(\mathrm{e}^{-\mathrm{x}}\)
2 \(\mathrm{e}^{-\mathrm{y}}\)
3 \(e^{\sin y}\)
4 \(e^{\cos y}\)
Explanation:
(A) : Given, \(\sin y\left(\frac{d y}{d x}\right)=\cos y(1-x \cos y)\) \(\frac{\sin y}{\cos y}\left(\frac{d y}{d x}\right)=1-x \cos y\) \(\tan y\left(\frac{d y}{d x}\right)=1-x \cos y\) \(\tan y\left(\frac{d y}{d x}\right)=1-\frac{x}{\sec y}\) \(\tan y \cdot \sec y\left(\frac{d y}{d x}\right)=\sec y-x\) Put, \(\quad \sec \mathrm{y}=\mathrm{t}\) \(\sec y \cdot \tan y\left(\frac{d y}{d x}\right)=\frac{d t}{d x}\) Then equation becomes, \(\frac{\mathrm{dt}}{\mathrm{dx}}-\mathrm{t}=-\mathrm{x}\) Which is the form of \(\frac{d y}{d x}+P y=Q\) \(\mathrm{P}=-1 \quad\) and \(\quad \mathrm{Q}=-\mathrm{X}\) I.F. \(=\mathrm{e}^{\int-1 \mathrm{dx}}=\mathrm{e}^{-\mathrm{x}}\)
MHT CET-2020
Differential Equation
87493
Solution of the differential equation \(\frac{d y}{d x}+2 y=e^{-x}\) is
87494
The integrating factor of the differential equation \( \frac{d y}{d x}+\frac{1}{x} y=x^{3}-3 \) is
1 -y
2 x
3 -x
4 y
Explanation:
(B) : Given, \(\frac{d y}{d x}+\frac{1}{x} y=x^{3}-3\) The above differential equation of the form \(\frac{d y}{d x}+P y=Q\) \(\mathrm{P}=\frac{1}{\mathrm{x}}\) and \(\mathrm{Q}=\mathrm{x}^{3}-3\) I. F. \(=\mathrm{e}^{\int P \mathrm{Pdx}}\) \(=\mathrm{e}^{\int \frac{1}{x} \mathrm{dx}}\) \(=\mathrm{e}^{\log \mathrm{x}}=\mathrm{x}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Differential Equation
87489
The differential \(y \frac{d y}{d x}+x=\) c represents
1 a family of parabolas
2 a family of circles whose centres are on the \(x\) axis
3 a family of hyperbolas
4 a family of circles whose centres are on the yaxis
Explanation:
(B) : Given, \[ \begin{array}{l} y \frac{d y}{d x}+x=c \\ y \frac{d y}{d x}=c-x \\ y \frac{d y}{d y}=(c-x) d x \end{array} \] On integration both sides, we get \[ \begin{array}{l} \int \mathrm{y} d \mathrm{~d}=\int(\mathrm{c}-\mathrm{x}) \mathrm{dx} \\ \frac{\mathrm{y}^{2}}{2}=\mathrm{cx}-\frac{\mathrm{x}^{2}}{2}+\mathrm{k} \\ \frac{\mathrm{x}^{2}}{2}+\frac{\mathrm{y}^{2}}{2}=\mathrm{cx}+\mathrm{k} \\ \mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{cx}-2 \mathrm{k}=0 \end{array} \] This equation is represent the circle whose centre are on the x -axis.
Karnataka CET-2008
Differential Equation
87490
The differential equation of the family of circles passing through the origin and having their centres on the \( x \)-axis is
1 \( x^{2}=y^{2}+x y \frac{d y}{d x} \)
2 \( x^{2}=y^{2}+3 x y \frac{d y}{d x} \)
3 \( y^{2}=x^{2}+2 x y \frac{d y}{d x} \)
4 \( y^{2}=x^{2}-2 x y \frac{d y}{d x} \)
Explanation:
(C) : Given that family circle passing through the origin and centre on \(\mathrm{x}\) - axis. Let centre be \((h, 0)\) and radius be \(h\). Now equation of family of circle, \((\mathrm{x}-\mathrm{h})^{2}+(\mathrm{y}-\mathrm{k})^{2}=\mathrm{h}^{2}\) \((x-h)^{2}+(y-0)^{2}=h^{2}\) \((\mathrm{x}-\mathrm{h})^{2}+\mathrm{y}^{2}=\mathrm{h}^{2}\) \(\mathrm{x}^{2}+\mathrm{h}^{2}-2 \mathrm{xh}+\mathrm{y}^{2}=\mathrm{h}^{2}\) \(\mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{xh}=0\) On differentiating both sides w.r.t.x, we get- \(2 x+2 y \frac{d y}{d x}-2 h=0\) \(x+y \frac{d y}{d x}=h\) On putting the value of \(h\) in equation (i) \(x^{2}+y^{2}-2 x\left(x+y \frac{d y}{d x}\right)=0\) \(\mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{x}^{2}-2 \mathrm{xy} \frac{\mathrm{dy}}{\mathrm{dx}}=0\) \(-x^{2}+y^{2}-2 x y \frac{d y}{d x}=0\) \(\mathrm{x}^{2}-\mathrm{y}^{2}+2 \mathrm{xy} \frac{\mathrm{dy}}{\mathrm{dx}}=0\) \(y^{2}=x^{2}+2 x y \frac{d y}{d x}\)
Karnataka CET-2009
Differential Equation
87491
The integrating factor of the differential equation \(\sin y\left(\frac{d y}{d x}\right)=\cos y(1-x \cos y)\) is
1 \(\mathrm{e}^{-\mathrm{x}}\)
2 \(\mathrm{e}^{-\mathrm{y}}\)
3 \(e^{\sin y}\)
4 \(e^{\cos y}\)
Explanation:
(A) : Given, \(\sin y\left(\frac{d y}{d x}\right)=\cos y(1-x \cos y)\) \(\frac{\sin y}{\cos y}\left(\frac{d y}{d x}\right)=1-x \cos y\) \(\tan y\left(\frac{d y}{d x}\right)=1-x \cos y\) \(\tan y\left(\frac{d y}{d x}\right)=1-\frac{x}{\sec y}\) \(\tan y \cdot \sec y\left(\frac{d y}{d x}\right)=\sec y-x\) Put, \(\quad \sec \mathrm{y}=\mathrm{t}\) \(\sec y \cdot \tan y\left(\frac{d y}{d x}\right)=\frac{d t}{d x}\) Then equation becomes, \(\frac{\mathrm{dt}}{\mathrm{dx}}-\mathrm{t}=-\mathrm{x}\) Which is the form of \(\frac{d y}{d x}+P y=Q\) \(\mathrm{P}=-1 \quad\) and \(\quad \mathrm{Q}=-\mathrm{X}\) I.F. \(=\mathrm{e}^{\int-1 \mathrm{dx}}=\mathrm{e}^{-\mathrm{x}}\)
MHT CET-2020
Differential Equation
87493
Solution of the differential equation \(\frac{d y}{d x}+2 y=e^{-x}\) is
87494
The integrating factor of the differential equation \( \frac{d y}{d x}+\frac{1}{x} y=x^{3}-3 \) is
1 -y
2 x
3 -x
4 y
Explanation:
(B) : Given, \(\frac{d y}{d x}+\frac{1}{x} y=x^{3}-3\) The above differential equation of the form \(\frac{d y}{d x}+P y=Q\) \(\mathrm{P}=\frac{1}{\mathrm{x}}\) and \(\mathrm{Q}=\mathrm{x}^{3}-3\) I. F. \(=\mathrm{e}^{\int P \mathrm{Pdx}}\) \(=\mathrm{e}^{\int \frac{1}{x} \mathrm{dx}}\) \(=\mathrm{e}^{\log \mathrm{x}}=\mathrm{x}\)
87489
The differential \(y \frac{d y}{d x}+x=\) c represents
1 a family of parabolas
2 a family of circles whose centres are on the \(x\) axis
3 a family of hyperbolas
4 a family of circles whose centres are on the yaxis
Explanation:
(B) : Given, \[ \begin{array}{l} y \frac{d y}{d x}+x=c \\ y \frac{d y}{d x}=c-x \\ y \frac{d y}{d y}=(c-x) d x \end{array} \] On integration both sides, we get \[ \begin{array}{l} \int \mathrm{y} d \mathrm{~d}=\int(\mathrm{c}-\mathrm{x}) \mathrm{dx} \\ \frac{\mathrm{y}^{2}}{2}=\mathrm{cx}-\frac{\mathrm{x}^{2}}{2}+\mathrm{k} \\ \frac{\mathrm{x}^{2}}{2}+\frac{\mathrm{y}^{2}}{2}=\mathrm{cx}+\mathrm{k} \\ \mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{cx}-2 \mathrm{k}=0 \end{array} \] This equation is represent the circle whose centre are on the x -axis.
Karnataka CET-2008
Differential Equation
87490
The differential equation of the family of circles passing through the origin and having their centres on the \( x \)-axis is
1 \( x^{2}=y^{2}+x y \frac{d y}{d x} \)
2 \( x^{2}=y^{2}+3 x y \frac{d y}{d x} \)
3 \( y^{2}=x^{2}+2 x y \frac{d y}{d x} \)
4 \( y^{2}=x^{2}-2 x y \frac{d y}{d x} \)
Explanation:
(C) : Given that family circle passing through the origin and centre on \(\mathrm{x}\) - axis. Let centre be \((h, 0)\) and radius be \(h\). Now equation of family of circle, \((\mathrm{x}-\mathrm{h})^{2}+(\mathrm{y}-\mathrm{k})^{2}=\mathrm{h}^{2}\) \((x-h)^{2}+(y-0)^{2}=h^{2}\) \((\mathrm{x}-\mathrm{h})^{2}+\mathrm{y}^{2}=\mathrm{h}^{2}\) \(\mathrm{x}^{2}+\mathrm{h}^{2}-2 \mathrm{xh}+\mathrm{y}^{2}=\mathrm{h}^{2}\) \(\mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{xh}=0\) On differentiating both sides w.r.t.x, we get- \(2 x+2 y \frac{d y}{d x}-2 h=0\) \(x+y \frac{d y}{d x}=h\) On putting the value of \(h\) in equation (i) \(x^{2}+y^{2}-2 x\left(x+y \frac{d y}{d x}\right)=0\) \(\mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{x}^{2}-2 \mathrm{xy} \frac{\mathrm{dy}}{\mathrm{dx}}=0\) \(-x^{2}+y^{2}-2 x y \frac{d y}{d x}=0\) \(\mathrm{x}^{2}-\mathrm{y}^{2}+2 \mathrm{xy} \frac{\mathrm{dy}}{\mathrm{dx}}=0\) \(y^{2}=x^{2}+2 x y \frac{d y}{d x}\)
Karnataka CET-2009
Differential Equation
87491
The integrating factor of the differential equation \(\sin y\left(\frac{d y}{d x}\right)=\cos y(1-x \cos y)\) is
1 \(\mathrm{e}^{-\mathrm{x}}\)
2 \(\mathrm{e}^{-\mathrm{y}}\)
3 \(e^{\sin y}\)
4 \(e^{\cos y}\)
Explanation:
(A) : Given, \(\sin y\left(\frac{d y}{d x}\right)=\cos y(1-x \cos y)\) \(\frac{\sin y}{\cos y}\left(\frac{d y}{d x}\right)=1-x \cos y\) \(\tan y\left(\frac{d y}{d x}\right)=1-x \cos y\) \(\tan y\left(\frac{d y}{d x}\right)=1-\frac{x}{\sec y}\) \(\tan y \cdot \sec y\left(\frac{d y}{d x}\right)=\sec y-x\) Put, \(\quad \sec \mathrm{y}=\mathrm{t}\) \(\sec y \cdot \tan y\left(\frac{d y}{d x}\right)=\frac{d t}{d x}\) Then equation becomes, \(\frac{\mathrm{dt}}{\mathrm{dx}}-\mathrm{t}=-\mathrm{x}\) Which is the form of \(\frac{d y}{d x}+P y=Q\) \(\mathrm{P}=-1 \quad\) and \(\quad \mathrm{Q}=-\mathrm{X}\) I.F. \(=\mathrm{e}^{\int-1 \mathrm{dx}}=\mathrm{e}^{-\mathrm{x}}\)
MHT CET-2020
Differential Equation
87493
Solution of the differential equation \(\frac{d y}{d x}+2 y=e^{-x}\) is
87494
The integrating factor of the differential equation \( \frac{d y}{d x}+\frac{1}{x} y=x^{3}-3 \) is
1 -y
2 x
3 -x
4 y
Explanation:
(B) : Given, \(\frac{d y}{d x}+\frac{1}{x} y=x^{3}-3\) The above differential equation of the form \(\frac{d y}{d x}+P y=Q\) \(\mathrm{P}=\frac{1}{\mathrm{x}}\) and \(\mathrm{Q}=\mathrm{x}^{3}-3\) I. F. \(=\mathrm{e}^{\int P \mathrm{Pdx}}\) \(=\mathrm{e}^{\int \frac{1}{x} \mathrm{dx}}\) \(=\mathrm{e}^{\log \mathrm{x}}=\mathrm{x}\)
87489
The differential \(y \frac{d y}{d x}+x=\) c represents
1 a family of parabolas
2 a family of circles whose centres are on the \(x\) axis
3 a family of hyperbolas
4 a family of circles whose centres are on the yaxis
Explanation:
(B) : Given, \[ \begin{array}{l} y \frac{d y}{d x}+x=c \\ y \frac{d y}{d x}=c-x \\ y \frac{d y}{d y}=(c-x) d x \end{array} \] On integration both sides, we get \[ \begin{array}{l} \int \mathrm{y} d \mathrm{~d}=\int(\mathrm{c}-\mathrm{x}) \mathrm{dx} \\ \frac{\mathrm{y}^{2}}{2}=\mathrm{cx}-\frac{\mathrm{x}^{2}}{2}+\mathrm{k} \\ \frac{\mathrm{x}^{2}}{2}+\frac{\mathrm{y}^{2}}{2}=\mathrm{cx}+\mathrm{k} \\ \mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{cx}-2 \mathrm{k}=0 \end{array} \] This equation is represent the circle whose centre are on the x -axis.
Karnataka CET-2008
Differential Equation
87490
The differential equation of the family of circles passing through the origin and having their centres on the \( x \)-axis is
1 \( x^{2}=y^{2}+x y \frac{d y}{d x} \)
2 \( x^{2}=y^{2}+3 x y \frac{d y}{d x} \)
3 \( y^{2}=x^{2}+2 x y \frac{d y}{d x} \)
4 \( y^{2}=x^{2}-2 x y \frac{d y}{d x} \)
Explanation:
(C) : Given that family circle passing through the origin and centre on \(\mathrm{x}\) - axis. Let centre be \((h, 0)\) and radius be \(h\). Now equation of family of circle, \((\mathrm{x}-\mathrm{h})^{2}+(\mathrm{y}-\mathrm{k})^{2}=\mathrm{h}^{2}\) \((x-h)^{2}+(y-0)^{2}=h^{2}\) \((\mathrm{x}-\mathrm{h})^{2}+\mathrm{y}^{2}=\mathrm{h}^{2}\) \(\mathrm{x}^{2}+\mathrm{h}^{2}-2 \mathrm{xh}+\mathrm{y}^{2}=\mathrm{h}^{2}\) \(\mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{xh}=0\) On differentiating both sides w.r.t.x, we get- \(2 x+2 y \frac{d y}{d x}-2 h=0\) \(x+y \frac{d y}{d x}=h\) On putting the value of \(h\) in equation (i) \(x^{2}+y^{2}-2 x\left(x+y \frac{d y}{d x}\right)=0\) \(\mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{x}^{2}-2 \mathrm{xy} \frac{\mathrm{dy}}{\mathrm{dx}}=0\) \(-x^{2}+y^{2}-2 x y \frac{d y}{d x}=0\) \(\mathrm{x}^{2}-\mathrm{y}^{2}+2 \mathrm{xy} \frac{\mathrm{dy}}{\mathrm{dx}}=0\) \(y^{2}=x^{2}+2 x y \frac{d y}{d x}\)
Karnataka CET-2009
Differential Equation
87491
The integrating factor of the differential equation \(\sin y\left(\frac{d y}{d x}\right)=\cos y(1-x \cos y)\) is
1 \(\mathrm{e}^{-\mathrm{x}}\)
2 \(\mathrm{e}^{-\mathrm{y}}\)
3 \(e^{\sin y}\)
4 \(e^{\cos y}\)
Explanation:
(A) : Given, \(\sin y\left(\frac{d y}{d x}\right)=\cos y(1-x \cos y)\) \(\frac{\sin y}{\cos y}\left(\frac{d y}{d x}\right)=1-x \cos y\) \(\tan y\left(\frac{d y}{d x}\right)=1-x \cos y\) \(\tan y\left(\frac{d y}{d x}\right)=1-\frac{x}{\sec y}\) \(\tan y \cdot \sec y\left(\frac{d y}{d x}\right)=\sec y-x\) Put, \(\quad \sec \mathrm{y}=\mathrm{t}\) \(\sec y \cdot \tan y\left(\frac{d y}{d x}\right)=\frac{d t}{d x}\) Then equation becomes, \(\frac{\mathrm{dt}}{\mathrm{dx}}-\mathrm{t}=-\mathrm{x}\) Which is the form of \(\frac{d y}{d x}+P y=Q\) \(\mathrm{P}=-1 \quad\) and \(\quad \mathrm{Q}=-\mathrm{X}\) I.F. \(=\mathrm{e}^{\int-1 \mathrm{dx}}=\mathrm{e}^{-\mathrm{x}}\)
MHT CET-2020
Differential Equation
87493
Solution of the differential equation \(\frac{d y}{d x}+2 y=e^{-x}\) is
87494
The integrating factor of the differential equation \( \frac{d y}{d x}+\frac{1}{x} y=x^{3}-3 \) is
1 -y
2 x
3 -x
4 y
Explanation:
(B) : Given, \(\frac{d y}{d x}+\frac{1}{x} y=x^{3}-3\) The above differential equation of the form \(\frac{d y}{d x}+P y=Q\) \(\mathrm{P}=\frac{1}{\mathrm{x}}\) and \(\mathrm{Q}=\mathrm{x}^{3}-3\) I. F. \(=\mathrm{e}^{\int P \mathrm{Pdx}}\) \(=\mathrm{e}^{\int \frac{1}{x} \mathrm{dx}}\) \(=\mathrm{e}^{\log \mathrm{x}}=\mathrm{x}\)