87262
The solution of the differential equation \(y-x \frac{d y}{d x}=a\left(y^{2}+\frac{d y}{d x}\right)\) is
1 \(y=C(x+a)(1-a y)\)
2 \(y=C(x+a)(1+a y)\)
3 \(y=C(x-a)(1-a y)\)
4 None of the above
Explanation:
(A) : Given, \(y-x \frac{d y}{d x}=a\left(y^{2}+\frac{d y}{d x}\right) \Rightarrow y-a y^{2}=(a+x) \frac{d y}{d x}\) \(\frac{d x}{a+x}=\frac{1}{y-a y^{2}} d y \Rightarrow \int \frac{1}{a+x} d x=\int \frac{1}{y-a y^{2}} d y\) \(\int \frac{1}{a+x} d x=\int \frac{1}{y(1-a y)} d y\) \(\int \frac{1}{a+x} d x=\int\left(\frac{1}{y}+\frac{a}{1-a y}\right) d y\) Let, \(\quad 1-\mathrm{ay}=\mathrm{t} \Rightarrow-\mathrm{ady}=\mathrm{dt}\) \(\int \frac{1}{a+x} d x=\int \frac{1}{y} d y-\int \frac{d t}{t}\) \(\log (a+x)+\log C=\log y-\log (1-a y)\) \(\log y=\log (x+a)+\log C+\log (1-a y)\) \(\operatorname{loy} y=\log [C(x+a)(1-a y)]\) \(y=C(x+a)(1-a y)\)
CG PET- 2016
Differential Equation
87263
The solution of differential equation \((2 y-1) d x-(2 x+3) d y=0\) will be
1 \(\frac{2 x-1}{2 y+3}=C\)
2 \(\frac{2 y+1}{2 x-3}=C\)
3 \(\frac{2 x+3}{2 y-1}=C\)
4 \(\frac{2 x-1}{2 y-1}=C\)
Explanation:
(C) : Given, Differential equation is, \((2 y-1) d x-(2 x+3) d y=0\) \((2 y-1) d x=(2 x+3) d y\) \(\left(\frac{1}{2 x+3}\right) d x=\left(\frac{1}{2 y-1}\right) d y\) \(\int \frac{2}{2 x+3} \mathrm{dx}=\int \frac{2}{2 \mathrm{y}-1} \mathrm{dy}\) \(\log (2 \mathrm{x}+3)=\log (2 \mathrm{y}-1)+\log \mathrm{C}\) \(\frac{2 \mathrm{x}+3}{2 \mathrm{y}-1}=\mathrm{C}\)
CG PET- 2016
Differential Equation
87264
The solution of the differential equation \(\frac{d y}{d x}=\frac{x \log x^{2}+x}{\sin y+y \cos y} \text { will be }\)
1 \(y \sin y=x^{2} \log x+C\)
2 \(y \sin y=x^{2}+C\)
3 \(y \sin y=x^{2}+\log x+C\)
4 \(y \sin y=x \log x+C\)
Explanation:
(A) : Given, \(\frac{d y}{d x}=\frac{x \log x^{2}+x}{\sin y+y \cos y}\) \((\sin y+y \cos y) d y=\left(x \log x^{2}+x\right) d x\) \((\sin y+y \cos y) d y=(2 x \log x+x) d x\) \(d(y \sin y)=d\left(x^{2} \log x\right)\) Integrating on both sides, \(y \sin y=x^{2} \log x+C\)
CG PET- 2016
Differential Equation
87265
If the solution of the differential equation \(\frac{d y}{d x}+e^{x}\left(x^{2}-2\right) y=\left(x^{2}-2 x\right)\left(x^{2}-2\right) e^{2 x}\) satisfies \(y(0)=0\), then the value of \(y(2)\) is
87262
The solution of the differential equation \(y-x \frac{d y}{d x}=a\left(y^{2}+\frac{d y}{d x}\right)\) is
1 \(y=C(x+a)(1-a y)\)
2 \(y=C(x+a)(1+a y)\)
3 \(y=C(x-a)(1-a y)\)
4 None of the above
Explanation:
(A) : Given, \(y-x \frac{d y}{d x}=a\left(y^{2}+\frac{d y}{d x}\right) \Rightarrow y-a y^{2}=(a+x) \frac{d y}{d x}\) \(\frac{d x}{a+x}=\frac{1}{y-a y^{2}} d y \Rightarrow \int \frac{1}{a+x} d x=\int \frac{1}{y-a y^{2}} d y\) \(\int \frac{1}{a+x} d x=\int \frac{1}{y(1-a y)} d y\) \(\int \frac{1}{a+x} d x=\int\left(\frac{1}{y}+\frac{a}{1-a y}\right) d y\) Let, \(\quad 1-\mathrm{ay}=\mathrm{t} \Rightarrow-\mathrm{ady}=\mathrm{dt}\) \(\int \frac{1}{a+x} d x=\int \frac{1}{y} d y-\int \frac{d t}{t}\) \(\log (a+x)+\log C=\log y-\log (1-a y)\) \(\log y=\log (x+a)+\log C+\log (1-a y)\) \(\operatorname{loy} y=\log [C(x+a)(1-a y)]\) \(y=C(x+a)(1-a y)\)
CG PET- 2016
Differential Equation
87263
The solution of differential equation \((2 y-1) d x-(2 x+3) d y=0\) will be
1 \(\frac{2 x-1}{2 y+3}=C\)
2 \(\frac{2 y+1}{2 x-3}=C\)
3 \(\frac{2 x+3}{2 y-1}=C\)
4 \(\frac{2 x-1}{2 y-1}=C\)
Explanation:
(C) : Given, Differential equation is, \((2 y-1) d x-(2 x+3) d y=0\) \((2 y-1) d x=(2 x+3) d y\) \(\left(\frac{1}{2 x+3}\right) d x=\left(\frac{1}{2 y-1}\right) d y\) \(\int \frac{2}{2 x+3} \mathrm{dx}=\int \frac{2}{2 \mathrm{y}-1} \mathrm{dy}\) \(\log (2 \mathrm{x}+3)=\log (2 \mathrm{y}-1)+\log \mathrm{C}\) \(\frac{2 \mathrm{x}+3}{2 \mathrm{y}-1}=\mathrm{C}\)
CG PET- 2016
Differential Equation
87264
The solution of the differential equation \(\frac{d y}{d x}=\frac{x \log x^{2}+x}{\sin y+y \cos y} \text { will be }\)
1 \(y \sin y=x^{2} \log x+C\)
2 \(y \sin y=x^{2}+C\)
3 \(y \sin y=x^{2}+\log x+C\)
4 \(y \sin y=x \log x+C\)
Explanation:
(A) : Given, \(\frac{d y}{d x}=\frac{x \log x^{2}+x}{\sin y+y \cos y}\) \((\sin y+y \cos y) d y=\left(x \log x^{2}+x\right) d x\) \((\sin y+y \cos y) d y=(2 x \log x+x) d x\) \(d(y \sin y)=d\left(x^{2} \log x\right)\) Integrating on both sides, \(y \sin y=x^{2} \log x+C\)
CG PET- 2016
Differential Equation
87265
If the solution of the differential equation \(\frac{d y}{d x}+e^{x}\left(x^{2}-2\right) y=\left(x^{2}-2 x\right)\left(x^{2}-2\right) e^{2 x}\) satisfies \(y(0)=0\), then the value of \(y(2)\) is
87262
The solution of the differential equation \(y-x \frac{d y}{d x}=a\left(y^{2}+\frac{d y}{d x}\right)\) is
1 \(y=C(x+a)(1-a y)\)
2 \(y=C(x+a)(1+a y)\)
3 \(y=C(x-a)(1-a y)\)
4 None of the above
Explanation:
(A) : Given, \(y-x \frac{d y}{d x}=a\left(y^{2}+\frac{d y}{d x}\right) \Rightarrow y-a y^{2}=(a+x) \frac{d y}{d x}\) \(\frac{d x}{a+x}=\frac{1}{y-a y^{2}} d y \Rightarrow \int \frac{1}{a+x} d x=\int \frac{1}{y-a y^{2}} d y\) \(\int \frac{1}{a+x} d x=\int \frac{1}{y(1-a y)} d y\) \(\int \frac{1}{a+x} d x=\int\left(\frac{1}{y}+\frac{a}{1-a y}\right) d y\) Let, \(\quad 1-\mathrm{ay}=\mathrm{t} \Rightarrow-\mathrm{ady}=\mathrm{dt}\) \(\int \frac{1}{a+x} d x=\int \frac{1}{y} d y-\int \frac{d t}{t}\) \(\log (a+x)+\log C=\log y-\log (1-a y)\) \(\log y=\log (x+a)+\log C+\log (1-a y)\) \(\operatorname{loy} y=\log [C(x+a)(1-a y)]\) \(y=C(x+a)(1-a y)\)
CG PET- 2016
Differential Equation
87263
The solution of differential equation \((2 y-1) d x-(2 x+3) d y=0\) will be
1 \(\frac{2 x-1}{2 y+3}=C\)
2 \(\frac{2 y+1}{2 x-3}=C\)
3 \(\frac{2 x+3}{2 y-1}=C\)
4 \(\frac{2 x-1}{2 y-1}=C\)
Explanation:
(C) : Given, Differential equation is, \((2 y-1) d x-(2 x+3) d y=0\) \((2 y-1) d x=(2 x+3) d y\) \(\left(\frac{1}{2 x+3}\right) d x=\left(\frac{1}{2 y-1}\right) d y\) \(\int \frac{2}{2 x+3} \mathrm{dx}=\int \frac{2}{2 \mathrm{y}-1} \mathrm{dy}\) \(\log (2 \mathrm{x}+3)=\log (2 \mathrm{y}-1)+\log \mathrm{C}\) \(\frac{2 \mathrm{x}+3}{2 \mathrm{y}-1}=\mathrm{C}\)
CG PET- 2016
Differential Equation
87264
The solution of the differential equation \(\frac{d y}{d x}=\frac{x \log x^{2}+x}{\sin y+y \cos y} \text { will be }\)
1 \(y \sin y=x^{2} \log x+C\)
2 \(y \sin y=x^{2}+C\)
3 \(y \sin y=x^{2}+\log x+C\)
4 \(y \sin y=x \log x+C\)
Explanation:
(A) : Given, \(\frac{d y}{d x}=\frac{x \log x^{2}+x}{\sin y+y \cos y}\) \((\sin y+y \cos y) d y=\left(x \log x^{2}+x\right) d x\) \((\sin y+y \cos y) d y=(2 x \log x+x) d x\) \(d(y \sin y)=d\left(x^{2} \log x\right)\) Integrating on both sides, \(y \sin y=x^{2} \log x+C\)
CG PET- 2016
Differential Equation
87265
If the solution of the differential equation \(\frac{d y}{d x}+e^{x}\left(x^{2}-2\right) y=\left(x^{2}-2 x\right)\left(x^{2}-2\right) e^{2 x}\) satisfies \(y(0)=0\), then the value of \(y(2)\) is
87262
The solution of the differential equation \(y-x \frac{d y}{d x}=a\left(y^{2}+\frac{d y}{d x}\right)\) is
1 \(y=C(x+a)(1-a y)\)
2 \(y=C(x+a)(1+a y)\)
3 \(y=C(x-a)(1-a y)\)
4 None of the above
Explanation:
(A) : Given, \(y-x \frac{d y}{d x}=a\left(y^{2}+\frac{d y}{d x}\right) \Rightarrow y-a y^{2}=(a+x) \frac{d y}{d x}\) \(\frac{d x}{a+x}=\frac{1}{y-a y^{2}} d y \Rightarrow \int \frac{1}{a+x} d x=\int \frac{1}{y-a y^{2}} d y\) \(\int \frac{1}{a+x} d x=\int \frac{1}{y(1-a y)} d y\) \(\int \frac{1}{a+x} d x=\int\left(\frac{1}{y}+\frac{a}{1-a y}\right) d y\) Let, \(\quad 1-\mathrm{ay}=\mathrm{t} \Rightarrow-\mathrm{ady}=\mathrm{dt}\) \(\int \frac{1}{a+x} d x=\int \frac{1}{y} d y-\int \frac{d t}{t}\) \(\log (a+x)+\log C=\log y-\log (1-a y)\) \(\log y=\log (x+a)+\log C+\log (1-a y)\) \(\operatorname{loy} y=\log [C(x+a)(1-a y)]\) \(y=C(x+a)(1-a y)\)
CG PET- 2016
Differential Equation
87263
The solution of differential equation \((2 y-1) d x-(2 x+3) d y=0\) will be
1 \(\frac{2 x-1}{2 y+3}=C\)
2 \(\frac{2 y+1}{2 x-3}=C\)
3 \(\frac{2 x+3}{2 y-1}=C\)
4 \(\frac{2 x-1}{2 y-1}=C\)
Explanation:
(C) : Given, Differential equation is, \((2 y-1) d x-(2 x+3) d y=0\) \((2 y-1) d x=(2 x+3) d y\) \(\left(\frac{1}{2 x+3}\right) d x=\left(\frac{1}{2 y-1}\right) d y\) \(\int \frac{2}{2 x+3} \mathrm{dx}=\int \frac{2}{2 \mathrm{y}-1} \mathrm{dy}\) \(\log (2 \mathrm{x}+3)=\log (2 \mathrm{y}-1)+\log \mathrm{C}\) \(\frac{2 \mathrm{x}+3}{2 \mathrm{y}-1}=\mathrm{C}\)
CG PET- 2016
Differential Equation
87264
The solution of the differential equation \(\frac{d y}{d x}=\frac{x \log x^{2}+x}{\sin y+y \cos y} \text { will be }\)
1 \(y \sin y=x^{2} \log x+C\)
2 \(y \sin y=x^{2}+C\)
3 \(y \sin y=x^{2}+\log x+C\)
4 \(y \sin y=x \log x+C\)
Explanation:
(A) : Given, \(\frac{d y}{d x}=\frac{x \log x^{2}+x}{\sin y+y \cos y}\) \((\sin y+y \cos y) d y=\left(x \log x^{2}+x\right) d x\) \((\sin y+y \cos y) d y=(2 x \log x+x) d x\) \(d(y \sin y)=d\left(x^{2} \log x\right)\) Integrating on both sides, \(y \sin y=x^{2} \log x+C\)
CG PET- 2016
Differential Equation
87265
If the solution of the differential equation \(\frac{d y}{d x}+e^{x}\left(x^{2}-2\right) y=\left(x^{2}-2 x\right)\left(x^{2}-2\right) e^{2 x}\) satisfies \(y(0)=0\), then the value of \(y(2)\) is
