(C) : Given, Differential equation, \(y=a e^{3 x}+b e^{x}\) Differentiating with respect to \(\mathrm{x}\), we get - \(\frac{d y}{d x}=3 a \cdot e^{3 x}+b e^{x}\) Again differentiating w.r. \(\mathrm{x}\), we get- \(\frac{d^{2} y}{d x^{2}}=9 a \cdot e^{3 x}+b e^{x}\) \(\frac{d^{2} y}{d x^{2}}=12 a e^{3 x}+4 b e^{x}-3 a e^{3 x}-3 b e^{x}\) \(\frac{d^{2} y}{d x^{2}}=4\left(a \cdot 3 e^{3 x}+b e^{x}\right)-3\left(a e^{3 x}+b e^{x}\right)\) \(\frac{d^{2} y}{d x^{2}}=4 \frac{d y}{d x}-3 y\) \(\frac{d^{2} y}{d x^{2}}-4 \frac{d y}{d x}+3 y=0\)
CG PET- 2011]**#
Differential Equation
87255
The differential equation of the family of circles touching the \(\mathbf{y}\)-axis at the origin is
1 \(x y^{\prime}-2 y=0\)
2 \(y^{\prime \prime}-4 y^{\prime}+4 y=0\)
3 \(2 x y y^{\prime}+x^{2}=y^{2}\)
4 \(2 y y^{\prime}+y^{2}=x^{2}\)
Explanation:
(C) : The equation of family of circle is, \((\mathrm{x}-\mathrm{h})^{2}+(\mathrm{y}-\mathrm{k})^{2}=\mathrm{r}^{2}\) Given, the system of circle touching \(y\) axis at the origin then centre \((\mathrm{h}, 0)\) and radius be \(\mathrm{h}\). Then equation of circle, \((x-h)^{2}+(y-0)^{2}=h^{2}\) \(x^{2}+h^{2}-2 x h+y^{2}=h^{2} \tag{i}\) \(x^{2}+y^{2}-2 x h=0\) Which is equation of circle touching y axis at origin. Differentiating with respect to \(x\), we get - \(2 x+2 y \frac{d y}{d x}-2 h=0 \Rightarrow h=x+y \frac{d y}{d x}\) Putting value of (h) in equation (i), we get - \(x^{2}+y^{2}-2 x\left(x+y \frac{d y}{d x}\right)=0\) \(x^{2}+y^{2}-2 x^{2}-2 x y \frac{d y}{d x}=0 \Rightarrow y^{2}-x^{2}-2 x y \frac{d y}{d x}=0\) \(2 x y \frac{d y}{d x}+x^{2}=y^{2} \Rightarrow 2 x y y^{\prime}+x^{2}=y^{2}\) \(\left(\because y^{\prime}=\frac{\mathrm{dy}}{\mathrm{dx}}\right)\)
CG PET- 2013
Differential Equation
87256
The solution of the differential equation \(x \frac{d y}{d x}=y-x \tan \left(\frac{y}{x}\right)\) is
1 \(x \sin \left(\frac{x}{y}\right)+C=0\)
2 \(x \sin y+C=0\)
3 \(x \sin \left(\frac{y}{x}\right)=C\)
4 None of the above
Explanation:
(C) : Given, Differential equation, \(x \frac{d y}{d x}=y-x \tan \left(\frac{y}{x}\right)\) \(\frac{d y}{d x}=\frac{y}{x}-\tan \left(\frac{y}{x}\right) \tag{i}\) Let, \(y=v x\) \(\frac{d y}{d x}=v+x \frac{d v}{d x}\) Then from equation (i), we get - \(v+x \frac{d v}{d x}=\frac{v x}{x}-\tan \left(\frac{v x}{x}\right)\) \(v+x \frac{d v}{d x}=v-\tan v\) \(x \frac{d v}{d x}=-\tan v \Rightarrow \quad \frac{d v}{\tan v}=-\frac{d x}{x}\) Integrating both side, we get - Let, \(\quad \sin \mathrm{v}=\mathrm{t}\) \(\int \cot v d v=\int-\frac{d x}{x} \Rightarrow \quad \int \frac{\cos v}{\sin v} d v=-\int \frac{d x}{x}\) \(\cos v=\frac{d t}{d v}\) \(\cos v d v=d t\) \(\therefore \quad \int \frac{1}{t} d t=-\int \frac{d x}{x}\) \(\log \mathrm{t}=-\log \mathrm{x}+\log \mathrm{C}\) \(\log \sin \mathrm{v}+\log \mathrm{x}=\log \mathrm{C}\) \(\log (\sin \mathrm{v} . \mathrm{x})=\log \mathrm{C}\) \(\mathrm{x} \sin \mathrm{v}=\mathrm{C}\) Putting \(y=v x\) or \(v=\frac{y}{x}\), we get - \(x \sin \left(\frac{y}{x}\right)=C\)
(C) : Given, Differential equation, \(y=a e^{3 x}+b e^{x}\) Differentiating with respect to \(\mathrm{x}\), we get - \(\frac{d y}{d x}=3 a \cdot e^{3 x}+b e^{x}\) Again differentiating w.r. \(\mathrm{x}\), we get- \(\frac{d^{2} y}{d x^{2}}=9 a \cdot e^{3 x}+b e^{x}\) \(\frac{d^{2} y}{d x^{2}}=12 a e^{3 x}+4 b e^{x}-3 a e^{3 x}-3 b e^{x}\) \(\frac{d^{2} y}{d x^{2}}=4\left(a \cdot 3 e^{3 x}+b e^{x}\right)-3\left(a e^{3 x}+b e^{x}\right)\) \(\frac{d^{2} y}{d x^{2}}=4 \frac{d y}{d x}-3 y\) \(\frac{d^{2} y}{d x^{2}}-4 \frac{d y}{d x}+3 y=0\)
CG PET- 2011]**#
Differential Equation
87255
The differential equation of the family of circles touching the \(\mathbf{y}\)-axis at the origin is
1 \(x y^{\prime}-2 y=0\)
2 \(y^{\prime \prime}-4 y^{\prime}+4 y=0\)
3 \(2 x y y^{\prime}+x^{2}=y^{2}\)
4 \(2 y y^{\prime}+y^{2}=x^{2}\)
Explanation:
(C) : The equation of family of circle is, \((\mathrm{x}-\mathrm{h})^{2}+(\mathrm{y}-\mathrm{k})^{2}=\mathrm{r}^{2}\) Given, the system of circle touching \(y\) axis at the origin then centre \((\mathrm{h}, 0)\) and radius be \(\mathrm{h}\). Then equation of circle, \((x-h)^{2}+(y-0)^{2}=h^{2}\) \(x^{2}+h^{2}-2 x h+y^{2}=h^{2} \tag{i}\) \(x^{2}+y^{2}-2 x h=0\) Which is equation of circle touching y axis at origin. Differentiating with respect to \(x\), we get - \(2 x+2 y \frac{d y}{d x}-2 h=0 \Rightarrow h=x+y \frac{d y}{d x}\) Putting value of (h) in equation (i), we get - \(x^{2}+y^{2}-2 x\left(x+y \frac{d y}{d x}\right)=0\) \(x^{2}+y^{2}-2 x^{2}-2 x y \frac{d y}{d x}=0 \Rightarrow y^{2}-x^{2}-2 x y \frac{d y}{d x}=0\) \(2 x y \frac{d y}{d x}+x^{2}=y^{2} \Rightarrow 2 x y y^{\prime}+x^{2}=y^{2}\) \(\left(\because y^{\prime}=\frac{\mathrm{dy}}{\mathrm{dx}}\right)\)
CG PET- 2013
Differential Equation
87256
The solution of the differential equation \(x \frac{d y}{d x}=y-x \tan \left(\frac{y}{x}\right)\) is
1 \(x \sin \left(\frac{x}{y}\right)+C=0\)
2 \(x \sin y+C=0\)
3 \(x \sin \left(\frac{y}{x}\right)=C\)
4 None of the above
Explanation:
(C) : Given, Differential equation, \(x \frac{d y}{d x}=y-x \tan \left(\frac{y}{x}\right)\) \(\frac{d y}{d x}=\frac{y}{x}-\tan \left(\frac{y}{x}\right) \tag{i}\) Let, \(y=v x\) \(\frac{d y}{d x}=v+x \frac{d v}{d x}\) Then from equation (i), we get - \(v+x \frac{d v}{d x}=\frac{v x}{x}-\tan \left(\frac{v x}{x}\right)\) \(v+x \frac{d v}{d x}=v-\tan v\) \(x \frac{d v}{d x}=-\tan v \Rightarrow \quad \frac{d v}{\tan v}=-\frac{d x}{x}\) Integrating both side, we get - Let, \(\quad \sin \mathrm{v}=\mathrm{t}\) \(\int \cot v d v=\int-\frac{d x}{x} \Rightarrow \quad \int \frac{\cos v}{\sin v} d v=-\int \frac{d x}{x}\) \(\cos v=\frac{d t}{d v}\) \(\cos v d v=d t\) \(\therefore \quad \int \frac{1}{t} d t=-\int \frac{d x}{x}\) \(\log \mathrm{t}=-\log \mathrm{x}+\log \mathrm{C}\) \(\log \sin \mathrm{v}+\log \mathrm{x}=\log \mathrm{C}\) \(\log (\sin \mathrm{v} . \mathrm{x})=\log \mathrm{C}\) \(\mathrm{x} \sin \mathrm{v}=\mathrm{C}\) Putting \(y=v x\) or \(v=\frac{y}{x}\), we get - \(x \sin \left(\frac{y}{x}\right)=C\)
(C) : Given, Differential equation, \(y=a e^{3 x}+b e^{x}\) Differentiating with respect to \(\mathrm{x}\), we get - \(\frac{d y}{d x}=3 a \cdot e^{3 x}+b e^{x}\) Again differentiating w.r. \(\mathrm{x}\), we get- \(\frac{d^{2} y}{d x^{2}}=9 a \cdot e^{3 x}+b e^{x}\) \(\frac{d^{2} y}{d x^{2}}=12 a e^{3 x}+4 b e^{x}-3 a e^{3 x}-3 b e^{x}\) \(\frac{d^{2} y}{d x^{2}}=4\left(a \cdot 3 e^{3 x}+b e^{x}\right)-3\left(a e^{3 x}+b e^{x}\right)\) \(\frac{d^{2} y}{d x^{2}}=4 \frac{d y}{d x}-3 y\) \(\frac{d^{2} y}{d x^{2}}-4 \frac{d y}{d x}+3 y=0\)
CG PET- 2011]**#
Differential Equation
87255
The differential equation of the family of circles touching the \(\mathbf{y}\)-axis at the origin is
1 \(x y^{\prime}-2 y=0\)
2 \(y^{\prime \prime}-4 y^{\prime}+4 y=0\)
3 \(2 x y y^{\prime}+x^{2}=y^{2}\)
4 \(2 y y^{\prime}+y^{2}=x^{2}\)
Explanation:
(C) : The equation of family of circle is, \((\mathrm{x}-\mathrm{h})^{2}+(\mathrm{y}-\mathrm{k})^{2}=\mathrm{r}^{2}\) Given, the system of circle touching \(y\) axis at the origin then centre \((\mathrm{h}, 0)\) and radius be \(\mathrm{h}\). Then equation of circle, \((x-h)^{2}+(y-0)^{2}=h^{2}\) \(x^{2}+h^{2}-2 x h+y^{2}=h^{2} \tag{i}\) \(x^{2}+y^{2}-2 x h=0\) Which is equation of circle touching y axis at origin. Differentiating with respect to \(x\), we get - \(2 x+2 y \frac{d y}{d x}-2 h=0 \Rightarrow h=x+y \frac{d y}{d x}\) Putting value of (h) in equation (i), we get - \(x^{2}+y^{2}-2 x\left(x+y \frac{d y}{d x}\right)=0\) \(x^{2}+y^{2}-2 x^{2}-2 x y \frac{d y}{d x}=0 \Rightarrow y^{2}-x^{2}-2 x y \frac{d y}{d x}=0\) \(2 x y \frac{d y}{d x}+x^{2}=y^{2} \Rightarrow 2 x y y^{\prime}+x^{2}=y^{2}\) \(\left(\because y^{\prime}=\frac{\mathrm{dy}}{\mathrm{dx}}\right)\)
CG PET- 2013
Differential Equation
87256
The solution of the differential equation \(x \frac{d y}{d x}=y-x \tan \left(\frac{y}{x}\right)\) is
1 \(x \sin \left(\frac{x}{y}\right)+C=0\)
2 \(x \sin y+C=0\)
3 \(x \sin \left(\frac{y}{x}\right)=C\)
4 None of the above
Explanation:
(C) : Given, Differential equation, \(x \frac{d y}{d x}=y-x \tan \left(\frac{y}{x}\right)\) \(\frac{d y}{d x}=\frac{y}{x}-\tan \left(\frac{y}{x}\right) \tag{i}\) Let, \(y=v x\) \(\frac{d y}{d x}=v+x \frac{d v}{d x}\) Then from equation (i), we get - \(v+x \frac{d v}{d x}=\frac{v x}{x}-\tan \left(\frac{v x}{x}\right)\) \(v+x \frac{d v}{d x}=v-\tan v\) \(x \frac{d v}{d x}=-\tan v \Rightarrow \quad \frac{d v}{\tan v}=-\frac{d x}{x}\) Integrating both side, we get - Let, \(\quad \sin \mathrm{v}=\mathrm{t}\) \(\int \cot v d v=\int-\frac{d x}{x} \Rightarrow \quad \int \frac{\cos v}{\sin v} d v=-\int \frac{d x}{x}\) \(\cos v=\frac{d t}{d v}\) \(\cos v d v=d t\) \(\therefore \quad \int \frac{1}{t} d t=-\int \frac{d x}{x}\) \(\log \mathrm{t}=-\log \mathrm{x}+\log \mathrm{C}\) \(\log \sin \mathrm{v}+\log \mathrm{x}=\log \mathrm{C}\) \(\log (\sin \mathrm{v} . \mathrm{x})=\log \mathrm{C}\) \(\mathrm{x} \sin \mathrm{v}=\mathrm{C}\) Putting \(y=v x\) or \(v=\frac{y}{x}\), we get - \(x \sin \left(\frac{y}{x}\right)=C\)
(C) : Given, Differential equation, \(y=a e^{3 x}+b e^{x}\) Differentiating with respect to \(\mathrm{x}\), we get - \(\frac{d y}{d x}=3 a \cdot e^{3 x}+b e^{x}\) Again differentiating w.r. \(\mathrm{x}\), we get- \(\frac{d^{2} y}{d x^{2}}=9 a \cdot e^{3 x}+b e^{x}\) \(\frac{d^{2} y}{d x^{2}}=12 a e^{3 x}+4 b e^{x}-3 a e^{3 x}-3 b e^{x}\) \(\frac{d^{2} y}{d x^{2}}=4\left(a \cdot 3 e^{3 x}+b e^{x}\right)-3\left(a e^{3 x}+b e^{x}\right)\) \(\frac{d^{2} y}{d x^{2}}=4 \frac{d y}{d x}-3 y\) \(\frac{d^{2} y}{d x^{2}}-4 \frac{d y}{d x}+3 y=0\)
CG PET- 2011]**#
Differential Equation
87255
The differential equation of the family of circles touching the \(\mathbf{y}\)-axis at the origin is
1 \(x y^{\prime}-2 y=0\)
2 \(y^{\prime \prime}-4 y^{\prime}+4 y=0\)
3 \(2 x y y^{\prime}+x^{2}=y^{2}\)
4 \(2 y y^{\prime}+y^{2}=x^{2}\)
Explanation:
(C) : The equation of family of circle is, \((\mathrm{x}-\mathrm{h})^{2}+(\mathrm{y}-\mathrm{k})^{2}=\mathrm{r}^{2}\) Given, the system of circle touching \(y\) axis at the origin then centre \((\mathrm{h}, 0)\) and radius be \(\mathrm{h}\). Then equation of circle, \((x-h)^{2}+(y-0)^{2}=h^{2}\) \(x^{2}+h^{2}-2 x h+y^{2}=h^{2} \tag{i}\) \(x^{2}+y^{2}-2 x h=0\) Which is equation of circle touching y axis at origin. Differentiating with respect to \(x\), we get - \(2 x+2 y \frac{d y}{d x}-2 h=0 \Rightarrow h=x+y \frac{d y}{d x}\) Putting value of (h) in equation (i), we get - \(x^{2}+y^{2}-2 x\left(x+y \frac{d y}{d x}\right)=0\) \(x^{2}+y^{2}-2 x^{2}-2 x y \frac{d y}{d x}=0 \Rightarrow y^{2}-x^{2}-2 x y \frac{d y}{d x}=0\) \(2 x y \frac{d y}{d x}+x^{2}=y^{2} \Rightarrow 2 x y y^{\prime}+x^{2}=y^{2}\) \(\left(\because y^{\prime}=\frac{\mathrm{dy}}{\mathrm{dx}}\right)\)
CG PET- 2013
Differential Equation
87256
The solution of the differential equation \(x \frac{d y}{d x}=y-x \tan \left(\frac{y}{x}\right)\) is
1 \(x \sin \left(\frac{x}{y}\right)+C=0\)
2 \(x \sin y+C=0\)
3 \(x \sin \left(\frac{y}{x}\right)=C\)
4 None of the above
Explanation:
(C) : Given, Differential equation, \(x \frac{d y}{d x}=y-x \tan \left(\frac{y}{x}\right)\) \(\frac{d y}{d x}=\frac{y}{x}-\tan \left(\frac{y}{x}\right) \tag{i}\) Let, \(y=v x\) \(\frac{d y}{d x}=v+x \frac{d v}{d x}\) Then from equation (i), we get - \(v+x \frac{d v}{d x}=\frac{v x}{x}-\tan \left(\frac{v x}{x}\right)\) \(v+x \frac{d v}{d x}=v-\tan v\) \(x \frac{d v}{d x}=-\tan v \Rightarrow \quad \frac{d v}{\tan v}=-\frac{d x}{x}\) Integrating both side, we get - Let, \(\quad \sin \mathrm{v}=\mathrm{t}\) \(\int \cot v d v=\int-\frac{d x}{x} \Rightarrow \quad \int \frac{\cos v}{\sin v} d v=-\int \frac{d x}{x}\) \(\cos v=\frac{d t}{d v}\) \(\cos v d v=d t\) \(\therefore \quad \int \frac{1}{t} d t=-\int \frac{d x}{x}\) \(\log \mathrm{t}=-\log \mathrm{x}+\log \mathrm{C}\) \(\log \sin \mathrm{v}+\log \mathrm{x}=\log \mathrm{C}\) \(\log (\sin \mathrm{v} . \mathrm{x})=\log \mathrm{C}\) \(\mathrm{x} \sin \mathrm{v}=\mathrm{C}\) Putting \(y=v x\) or \(v=\frac{y}{x}\), we get - \(x \sin \left(\frac{y}{x}\right)=C\)