87499
The integrating factor of \(\left(y e^{y}\right) d x=\left(y^{3}+2 x e^{y}\right) d y\) is
1 \(\frac{1}{y^{2}}\)
2 \(y^{2}\)
3 \(\frac{1}{\mathrm{x}^{2}}\)
4 \(x^{2}\)
Explanation:
(A) : Given, \(\mathrm{ye}^{\mathrm{y}} \mathrm{dx}=\left(\mathrm{y}^{3}+2 \mathrm{xe}^{\mathrm{y}}\right) \mathrm{dy}\) \(\frac{d x}{d y}=\frac{y^{3}+2 e^{y} x}{y^{y}}\) \(\frac{d x}{d y}=y^{2} e^{-y}+\frac{2 x}{y}\) \(\frac{d x}{d y}-2 \frac{x}{y}=y^{2} e^{-y}\) Now, which is form of \(\frac{d y}{d x}+P y=Q\) \(\mathrm{P}=\frac{-2}{\mathrm{y}}, \mathrm{Q}=\mathrm{y}^{2} \mathrm{e}^{-\mathrm{y}}\) \(I . F=e^{\int \text { Pdy }}=e^{\int-\frac{2}{y} d y}=e^{-2 \log y}=e^{\log \left(\frac{1}{y^{2}}\right)}=\frac{1}{y^{2}}\)
MHT CET-2019
Differential Equation
87500
The integrating factor of the differential equation \(\left(1+y^{2}\right) d x+\left(x-e^{-\tan ^{-1} y}\right) d y=0\)
1 \(\mathrm{e}^{\tan ^{-1} y}\)
2 \(\mathrm{ye}^{\tan ^{-1} y}\)
3 \(-\mathrm{ye}^{\tan ^{-1} y}\)
4 \(\mathrm{xe}^{\tan ^{-1} \mathrm{y}}\)
Explanation:
(A) : Given, differential equation \(\left(1+y^{2}\right) d x+\left(x-e^{-\tan ^{-1} y}\right) d y=0\) \(\frac{d x}{d y}=-\frac{\left(x-e^{-\tan ^{-1} y}\right)}{1+y^{2}} \Rightarrow \frac{d x}{d y}+\frac{x}{1+y^{2}}=\frac{e^{\tan ^{-1} y}}{1+y^{2}}\) Which is form of \(\frac{d y}{d x}+P x=Q\) I.F. \(=\mathrm{e}^{\int \mathrm{Pdy}}=\mathrm{e}^{\int \frac{1}{1+\mathrm{y}^{2}} \mathrm{dy}}=\mathrm{e}^{\tan ^{-1} \mathrm{y}}\)
MHT CET-2019
Differential Equation
87501
If \(\sin x\) is integrating factor (I.F.) of the linear differential equation \(\frac{d y}{d x}+P y=Q\), then \(P\) is
1 \(\log \sin x\)
2 \(\cos x\)
3 \(\tan x\)
4 \(\cot x\)
Explanation:
(D) : We have given that \(\sin \mathrm{x}\) is integrating factor of the linear differential equation \(e^{\int \mathrm{Pdx}=\sin \mathrm{x}}\) Taking \(\log\) on both side \(\log \left(\mathrm{e}^{\int \mathrm{Pdx}}\right)=\log (\sin \mathrm{x})\) \(\int \mathrm{Pdx} \log \mathrm{e}=\log \sin \mathrm{x}\) \(\int \mathrm{Pdx}=\log \sin \mathrm{x}\) Now differentiating w.r.t. \(\mathrm{x}\), we get- \(\frac{\mathrm{d}}{\mathrm{dx}}\left(\int \mathrm{Pdx}\right)=\frac{\mathrm{d}}{\mathrm{dx}} \log \sin \mathrm{x}\) \(P=\frac{1}{\sin x} \cos x\) \(\mathrm{P}=\cot \mathrm{x}\)
JCECE-2010
Differential Equation
87502
The integrating factor of the differential equation \(\frac{d y}{d x}(x \log x)+y=2 \log x\) is given by
1 \(e^{\mathrm{x}}\)
2 \(\log x\)
3 \(\log (\log x)\)
4 \(x\)
Explanation:
(B) : Given, differential equation \[ \begin{array}{l} \frac{d y}{d x}(x \log x)+y=2 \log x \\ \frac{d y}{d x}+\frac{y}{x \log x}=\frac{2}{x} \end{array} \] Which is the form of \( \frac{d y}{d x}+P y=Q \) \[ \text { I.F }=\mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\int \frac{1}{\mathrm{x} \log \mathrm{x}} \mathrm{dx}}=\mathrm{e}^{\log (\log \mathrm{x})}=\log \mathrm{x} \]
87499
The integrating factor of \(\left(y e^{y}\right) d x=\left(y^{3}+2 x e^{y}\right) d y\) is
1 \(\frac{1}{y^{2}}\)
2 \(y^{2}\)
3 \(\frac{1}{\mathrm{x}^{2}}\)
4 \(x^{2}\)
Explanation:
(A) : Given, \(\mathrm{ye}^{\mathrm{y}} \mathrm{dx}=\left(\mathrm{y}^{3}+2 \mathrm{xe}^{\mathrm{y}}\right) \mathrm{dy}\) \(\frac{d x}{d y}=\frac{y^{3}+2 e^{y} x}{y^{y}}\) \(\frac{d x}{d y}=y^{2} e^{-y}+\frac{2 x}{y}\) \(\frac{d x}{d y}-2 \frac{x}{y}=y^{2} e^{-y}\) Now, which is form of \(\frac{d y}{d x}+P y=Q\) \(\mathrm{P}=\frac{-2}{\mathrm{y}}, \mathrm{Q}=\mathrm{y}^{2} \mathrm{e}^{-\mathrm{y}}\) \(I . F=e^{\int \text { Pdy }}=e^{\int-\frac{2}{y} d y}=e^{-2 \log y}=e^{\log \left(\frac{1}{y^{2}}\right)}=\frac{1}{y^{2}}\)
MHT CET-2019
Differential Equation
87500
The integrating factor of the differential equation \(\left(1+y^{2}\right) d x+\left(x-e^{-\tan ^{-1} y}\right) d y=0\)
1 \(\mathrm{e}^{\tan ^{-1} y}\)
2 \(\mathrm{ye}^{\tan ^{-1} y}\)
3 \(-\mathrm{ye}^{\tan ^{-1} y}\)
4 \(\mathrm{xe}^{\tan ^{-1} \mathrm{y}}\)
Explanation:
(A) : Given, differential equation \(\left(1+y^{2}\right) d x+\left(x-e^{-\tan ^{-1} y}\right) d y=0\) \(\frac{d x}{d y}=-\frac{\left(x-e^{-\tan ^{-1} y}\right)}{1+y^{2}} \Rightarrow \frac{d x}{d y}+\frac{x}{1+y^{2}}=\frac{e^{\tan ^{-1} y}}{1+y^{2}}\) Which is form of \(\frac{d y}{d x}+P x=Q\) I.F. \(=\mathrm{e}^{\int \mathrm{Pdy}}=\mathrm{e}^{\int \frac{1}{1+\mathrm{y}^{2}} \mathrm{dy}}=\mathrm{e}^{\tan ^{-1} \mathrm{y}}\)
MHT CET-2019
Differential Equation
87501
If \(\sin x\) is integrating factor (I.F.) of the linear differential equation \(\frac{d y}{d x}+P y=Q\), then \(P\) is
1 \(\log \sin x\)
2 \(\cos x\)
3 \(\tan x\)
4 \(\cot x\)
Explanation:
(D) : We have given that \(\sin \mathrm{x}\) is integrating factor of the linear differential equation \(e^{\int \mathrm{Pdx}=\sin \mathrm{x}}\) Taking \(\log\) on both side \(\log \left(\mathrm{e}^{\int \mathrm{Pdx}}\right)=\log (\sin \mathrm{x})\) \(\int \mathrm{Pdx} \log \mathrm{e}=\log \sin \mathrm{x}\) \(\int \mathrm{Pdx}=\log \sin \mathrm{x}\) Now differentiating w.r.t. \(\mathrm{x}\), we get- \(\frac{\mathrm{d}}{\mathrm{dx}}\left(\int \mathrm{Pdx}\right)=\frac{\mathrm{d}}{\mathrm{dx}} \log \sin \mathrm{x}\) \(P=\frac{1}{\sin x} \cos x\) \(\mathrm{P}=\cot \mathrm{x}\)
JCECE-2010
Differential Equation
87502
The integrating factor of the differential equation \(\frac{d y}{d x}(x \log x)+y=2 \log x\) is given by
1 \(e^{\mathrm{x}}\)
2 \(\log x\)
3 \(\log (\log x)\)
4 \(x\)
Explanation:
(B) : Given, differential equation \[ \begin{array}{l} \frac{d y}{d x}(x \log x)+y=2 \log x \\ \frac{d y}{d x}+\frac{y}{x \log x}=\frac{2}{x} \end{array} \] Which is the form of \( \frac{d y}{d x}+P y=Q \) \[ \text { I.F }=\mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\int \frac{1}{\mathrm{x} \log \mathrm{x}} \mathrm{dx}}=\mathrm{e}^{\log (\log \mathrm{x})}=\log \mathrm{x} \]
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Differential Equation
87499
The integrating factor of \(\left(y e^{y}\right) d x=\left(y^{3}+2 x e^{y}\right) d y\) is
1 \(\frac{1}{y^{2}}\)
2 \(y^{2}\)
3 \(\frac{1}{\mathrm{x}^{2}}\)
4 \(x^{2}\)
Explanation:
(A) : Given, \(\mathrm{ye}^{\mathrm{y}} \mathrm{dx}=\left(\mathrm{y}^{3}+2 \mathrm{xe}^{\mathrm{y}}\right) \mathrm{dy}\) \(\frac{d x}{d y}=\frac{y^{3}+2 e^{y} x}{y^{y}}\) \(\frac{d x}{d y}=y^{2} e^{-y}+\frac{2 x}{y}\) \(\frac{d x}{d y}-2 \frac{x}{y}=y^{2} e^{-y}\) Now, which is form of \(\frac{d y}{d x}+P y=Q\) \(\mathrm{P}=\frac{-2}{\mathrm{y}}, \mathrm{Q}=\mathrm{y}^{2} \mathrm{e}^{-\mathrm{y}}\) \(I . F=e^{\int \text { Pdy }}=e^{\int-\frac{2}{y} d y}=e^{-2 \log y}=e^{\log \left(\frac{1}{y^{2}}\right)}=\frac{1}{y^{2}}\)
MHT CET-2019
Differential Equation
87500
The integrating factor of the differential equation \(\left(1+y^{2}\right) d x+\left(x-e^{-\tan ^{-1} y}\right) d y=0\)
1 \(\mathrm{e}^{\tan ^{-1} y}\)
2 \(\mathrm{ye}^{\tan ^{-1} y}\)
3 \(-\mathrm{ye}^{\tan ^{-1} y}\)
4 \(\mathrm{xe}^{\tan ^{-1} \mathrm{y}}\)
Explanation:
(A) : Given, differential equation \(\left(1+y^{2}\right) d x+\left(x-e^{-\tan ^{-1} y}\right) d y=0\) \(\frac{d x}{d y}=-\frac{\left(x-e^{-\tan ^{-1} y}\right)}{1+y^{2}} \Rightarrow \frac{d x}{d y}+\frac{x}{1+y^{2}}=\frac{e^{\tan ^{-1} y}}{1+y^{2}}\) Which is form of \(\frac{d y}{d x}+P x=Q\) I.F. \(=\mathrm{e}^{\int \mathrm{Pdy}}=\mathrm{e}^{\int \frac{1}{1+\mathrm{y}^{2}} \mathrm{dy}}=\mathrm{e}^{\tan ^{-1} \mathrm{y}}\)
MHT CET-2019
Differential Equation
87501
If \(\sin x\) is integrating factor (I.F.) of the linear differential equation \(\frac{d y}{d x}+P y=Q\), then \(P\) is
1 \(\log \sin x\)
2 \(\cos x\)
3 \(\tan x\)
4 \(\cot x\)
Explanation:
(D) : We have given that \(\sin \mathrm{x}\) is integrating factor of the linear differential equation \(e^{\int \mathrm{Pdx}=\sin \mathrm{x}}\) Taking \(\log\) on both side \(\log \left(\mathrm{e}^{\int \mathrm{Pdx}}\right)=\log (\sin \mathrm{x})\) \(\int \mathrm{Pdx} \log \mathrm{e}=\log \sin \mathrm{x}\) \(\int \mathrm{Pdx}=\log \sin \mathrm{x}\) Now differentiating w.r.t. \(\mathrm{x}\), we get- \(\frac{\mathrm{d}}{\mathrm{dx}}\left(\int \mathrm{Pdx}\right)=\frac{\mathrm{d}}{\mathrm{dx}} \log \sin \mathrm{x}\) \(P=\frac{1}{\sin x} \cos x\) \(\mathrm{P}=\cot \mathrm{x}\)
JCECE-2010
Differential Equation
87502
The integrating factor of the differential equation \(\frac{d y}{d x}(x \log x)+y=2 \log x\) is given by
1 \(e^{\mathrm{x}}\)
2 \(\log x\)
3 \(\log (\log x)\)
4 \(x\)
Explanation:
(B) : Given, differential equation \[ \begin{array}{l} \frac{d y}{d x}(x \log x)+y=2 \log x \\ \frac{d y}{d x}+\frac{y}{x \log x}=\frac{2}{x} \end{array} \] Which is the form of \( \frac{d y}{d x}+P y=Q \) \[ \text { I.F }=\mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\int \frac{1}{\mathrm{x} \log \mathrm{x}} \mathrm{dx}}=\mathrm{e}^{\log (\log \mathrm{x})}=\log \mathrm{x} \]
87499
The integrating factor of \(\left(y e^{y}\right) d x=\left(y^{3}+2 x e^{y}\right) d y\) is
1 \(\frac{1}{y^{2}}\)
2 \(y^{2}\)
3 \(\frac{1}{\mathrm{x}^{2}}\)
4 \(x^{2}\)
Explanation:
(A) : Given, \(\mathrm{ye}^{\mathrm{y}} \mathrm{dx}=\left(\mathrm{y}^{3}+2 \mathrm{xe}^{\mathrm{y}}\right) \mathrm{dy}\) \(\frac{d x}{d y}=\frac{y^{3}+2 e^{y} x}{y^{y}}\) \(\frac{d x}{d y}=y^{2} e^{-y}+\frac{2 x}{y}\) \(\frac{d x}{d y}-2 \frac{x}{y}=y^{2} e^{-y}\) Now, which is form of \(\frac{d y}{d x}+P y=Q\) \(\mathrm{P}=\frac{-2}{\mathrm{y}}, \mathrm{Q}=\mathrm{y}^{2} \mathrm{e}^{-\mathrm{y}}\) \(I . F=e^{\int \text { Pdy }}=e^{\int-\frac{2}{y} d y}=e^{-2 \log y}=e^{\log \left(\frac{1}{y^{2}}\right)}=\frac{1}{y^{2}}\)
MHT CET-2019
Differential Equation
87500
The integrating factor of the differential equation \(\left(1+y^{2}\right) d x+\left(x-e^{-\tan ^{-1} y}\right) d y=0\)
1 \(\mathrm{e}^{\tan ^{-1} y}\)
2 \(\mathrm{ye}^{\tan ^{-1} y}\)
3 \(-\mathrm{ye}^{\tan ^{-1} y}\)
4 \(\mathrm{xe}^{\tan ^{-1} \mathrm{y}}\)
Explanation:
(A) : Given, differential equation \(\left(1+y^{2}\right) d x+\left(x-e^{-\tan ^{-1} y}\right) d y=0\) \(\frac{d x}{d y}=-\frac{\left(x-e^{-\tan ^{-1} y}\right)}{1+y^{2}} \Rightarrow \frac{d x}{d y}+\frac{x}{1+y^{2}}=\frac{e^{\tan ^{-1} y}}{1+y^{2}}\) Which is form of \(\frac{d y}{d x}+P x=Q\) I.F. \(=\mathrm{e}^{\int \mathrm{Pdy}}=\mathrm{e}^{\int \frac{1}{1+\mathrm{y}^{2}} \mathrm{dy}}=\mathrm{e}^{\tan ^{-1} \mathrm{y}}\)
MHT CET-2019
Differential Equation
87501
If \(\sin x\) is integrating factor (I.F.) of the linear differential equation \(\frac{d y}{d x}+P y=Q\), then \(P\) is
1 \(\log \sin x\)
2 \(\cos x\)
3 \(\tan x\)
4 \(\cot x\)
Explanation:
(D) : We have given that \(\sin \mathrm{x}\) is integrating factor of the linear differential equation \(e^{\int \mathrm{Pdx}=\sin \mathrm{x}}\) Taking \(\log\) on both side \(\log \left(\mathrm{e}^{\int \mathrm{Pdx}}\right)=\log (\sin \mathrm{x})\) \(\int \mathrm{Pdx} \log \mathrm{e}=\log \sin \mathrm{x}\) \(\int \mathrm{Pdx}=\log \sin \mathrm{x}\) Now differentiating w.r.t. \(\mathrm{x}\), we get- \(\frac{\mathrm{d}}{\mathrm{dx}}\left(\int \mathrm{Pdx}\right)=\frac{\mathrm{d}}{\mathrm{dx}} \log \sin \mathrm{x}\) \(P=\frac{1}{\sin x} \cos x\) \(\mathrm{P}=\cot \mathrm{x}\)
JCECE-2010
Differential Equation
87502
The integrating factor of the differential equation \(\frac{d y}{d x}(x \log x)+y=2 \log x\) is given by
1 \(e^{\mathrm{x}}\)
2 \(\log x\)
3 \(\log (\log x)\)
4 \(x\)
Explanation:
(B) : Given, differential equation \[ \begin{array}{l} \frac{d y}{d x}(x \log x)+y=2 \log x \\ \frac{d y}{d x}+\frac{y}{x \log x}=\frac{2}{x} \end{array} \] Which is the form of \( \frac{d y}{d x}+P y=Q \) \[ \text { I.F }=\mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\int \frac{1}{\mathrm{x} \log \mathrm{x}} \mathrm{dx}}=\mathrm{e}^{\log (\log \mathrm{x})}=\log \mathrm{x} \]