(D) : Given, \(\mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y} \log \mathrm{x}=\mathrm{x}^{2}\) \(\frac{d y}{d x}+\frac{y \log x}{x}=x\) Now, which is the form \(\frac{d y}{d x}+P y=Q\) \(\mathrm{P}=\frac{\log \mathrm{x}}{\mathrm{x}}\) and \(\mathrm{Q}=\mathrm{x}\) \(I . F=e^{\int P d x}=e^{\int \frac{\log x}{x} d x}\) I.F. \(=\mathrm{e}^{\frac{(\log \mathrm{x})^{2}}{2}}\)
MHT CET-2020
Differential Equation
87498
The integrating factor of differential equation \(\left(1+y+x^{2} y\right) d x+\left(x+x^{3}\right) d y=0\)
1 \(x\)
2 \(\frac{1}{\mathrm{x}}\)
3 \(\log \mathrm{x}\)
4 \(e^{x}\)
Explanation:
(A) : Given, \(\left(1+y+x^{2} y\right) d x+\left(x+x^{3}\right) d y=0\) \(\frac{d y}{d x}=-\frac{\left(1+y+x^{2} y\right)}{\left(x+x^{3}\right)}\) \(\frac{d y}{d x}+\frac{1+y+x^{2} y}{x+x^{3}}=0\) \(\frac{d y}{d x}+\frac{1+y\left(1+x^{2}\right)}{x\left(1+x^{2}\right)}=0\) \(\frac{d y}{d x}+\frac{y}{x}+\frac{1}{x\left(1+x^{2}\right)}=0\) \(\frac{d y}{d x}+\frac{y}{x}=-\frac{1}{x}\left(1+x^{2}\right)\) It is the form of \(\frac{d y}{d x}+P y=Q\) \(\mathrm{P}=\frac{1}{\mathrm{x}} \quad\) and \(\mathrm{Q}=-\frac{1}{\mathrm{x}\left(1+\mathrm{x}^{2}\right)}\) Now, integrating factor I.F. \(=\mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\int \frac{1}{\mathrm{x} x}}=\mathrm{e}^{\log \mathrm{x}}\) I.F. \(=\mathrm{x}\)
(D) : Given, \(\mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y} \log \mathrm{x}=\mathrm{x}^{2}\) \(\frac{d y}{d x}+\frac{y \log x}{x}=x\) Now, which is the form \(\frac{d y}{d x}+P y=Q\) \(\mathrm{P}=\frac{\log \mathrm{x}}{\mathrm{x}}\) and \(\mathrm{Q}=\mathrm{x}\) \(I . F=e^{\int P d x}=e^{\int \frac{\log x}{x} d x}\) I.F. \(=\mathrm{e}^{\frac{(\log \mathrm{x})^{2}}{2}}\)
MHT CET-2020
Differential Equation
87498
The integrating factor of differential equation \(\left(1+y+x^{2} y\right) d x+\left(x+x^{3}\right) d y=0\)
1 \(x\)
2 \(\frac{1}{\mathrm{x}}\)
3 \(\log \mathrm{x}\)
4 \(e^{x}\)
Explanation:
(A) : Given, \(\left(1+y+x^{2} y\right) d x+\left(x+x^{3}\right) d y=0\) \(\frac{d y}{d x}=-\frac{\left(1+y+x^{2} y\right)}{\left(x+x^{3}\right)}\) \(\frac{d y}{d x}+\frac{1+y+x^{2} y}{x+x^{3}}=0\) \(\frac{d y}{d x}+\frac{1+y\left(1+x^{2}\right)}{x\left(1+x^{2}\right)}=0\) \(\frac{d y}{d x}+\frac{y}{x}+\frac{1}{x\left(1+x^{2}\right)}=0\) \(\frac{d y}{d x}+\frac{y}{x}=-\frac{1}{x}\left(1+x^{2}\right)\) It is the form of \(\frac{d y}{d x}+P y=Q\) \(\mathrm{P}=\frac{1}{\mathrm{x}} \quad\) and \(\mathrm{Q}=-\frac{1}{\mathrm{x}\left(1+\mathrm{x}^{2}\right)}\) Now, integrating factor I.F. \(=\mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\int \frac{1}{\mathrm{x} x}}=\mathrm{e}^{\log \mathrm{x}}\) I.F. \(=\mathrm{x}\)
(D) : Given, \(\mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y} \log \mathrm{x}=\mathrm{x}^{2}\) \(\frac{d y}{d x}+\frac{y \log x}{x}=x\) Now, which is the form \(\frac{d y}{d x}+P y=Q\) \(\mathrm{P}=\frac{\log \mathrm{x}}{\mathrm{x}}\) and \(\mathrm{Q}=\mathrm{x}\) \(I . F=e^{\int P d x}=e^{\int \frac{\log x}{x} d x}\) I.F. \(=\mathrm{e}^{\frac{(\log \mathrm{x})^{2}}{2}}\)
MHT CET-2020
Differential Equation
87498
The integrating factor of differential equation \(\left(1+y+x^{2} y\right) d x+\left(x+x^{3}\right) d y=0\)
1 \(x\)
2 \(\frac{1}{\mathrm{x}}\)
3 \(\log \mathrm{x}\)
4 \(e^{x}\)
Explanation:
(A) : Given, \(\left(1+y+x^{2} y\right) d x+\left(x+x^{3}\right) d y=0\) \(\frac{d y}{d x}=-\frac{\left(1+y+x^{2} y\right)}{\left(x+x^{3}\right)}\) \(\frac{d y}{d x}+\frac{1+y+x^{2} y}{x+x^{3}}=0\) \(\frac{d y}{d x}+\frac{1+y\left(1+x^{2}\right)}{x\left(1+x^{2}\right)}=0\) \(\frac{d y}{d x}+\frac{y}{x}+\frac{1}{x\left(1+x^{2}\right)}=0\) \(\frac{d y}{d x}+\frac{y}{x}=-\frac{1}{x}\left(1+x^{2}\right)\) It is the form of \(\frac{d y}{d x}+P y=Q\) \(\mathrm{P}=\frac{1}{\mathrm{x}} \quad\) and \(\mathrm{Q}=-\frac{1}{\mathrm{x}\left(1+\mathrm{x}^{2}\right)}\) Now, integrating factor I.F. \(=\mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\int \frac{1}{\mathrm{x} x}}=\mathrm{e}^{\log \mathrm{x}}\) I.F. \(=\mathrm{x}\)
(D) : Given, \(\mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y} \log \mathrm{x}=\mathrm{x}^{2}\) \(\frac{d y}{d x}+\frac{y \log x}{x}=x\) Now, which is the form \(\frac{d y}{d x}+P y=Q\) \(\mathrm{P}=\frac{\log \mathrm{x}}{\mathrm{x}}\) and \(\mathrm{Q}=\mathrm{x}\) \(I . F=e^{\int P d x}=e^{\int \frac{\log x}{x} d x}\) I.F. \(=\mathrm{e}^{\frac{(\log \mathrm{x})^{2}}{2}}\)
MHT CET-2020
Differential Equation
87498
The integrating factor of differential equation \(\left(1+y+x^{2} y\right) d x+\left(x+x^{3}\right) d y=0\)
1 \(x\)
2 \(\frac{1}{\mathrm{x}}\)
3 \(\log \mathrm{x}\)
4 \(e^{x}\)
Explanation:
(A) : Given, \(\left(1+y+x^{2} y\right) d x+\left(x+x^{3}\right) d y=0\) \(\frac{d y}{d x}=-\frac{\left(1+y+x^{2} y\right)}{\left(x+x^{3}\right)}\) \(\frac{d y}{d x}+\frac{1+y+x^{2} y}{x+x^{3}}=0\) \(\frac{d y}{d x}+\frac{1+y\left(1+x^{2}\right)}{x\left(1+x^{2}\right)}=0\) \(\frac{d y}{d x}+\frac{y}{x}+\frac{1}{x\left(1+x^{2}\right)}=0\) \(\frac{d y}{d x}+\frac{y}{x}=-\frac{1}{x}\left(1+x^{2}\right)\) It is the form of \(\frac{d y}{d x}+P y=Q\) \(\mathrm{P}=\frac{1}{\mathrm{x}} \quad\) and \(\mathrm{Q}=-\frac{1}{\mathrm{x}\left(1+\mathrm{x}^{2}\right)}\) Now, integrating factor I.F. \(=\mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\int \frac{1}{\mathrm{x} x}}=\mathrm{e}^{\log \mathrm{x}}\) I.F. \(=\mathrm{x}\)
