(C) : The given differential equation can be written as- \(x=e^{x y \frac{d y}{d x}}\) \(\log x=x y \frac{d y}{d x}\) \(y d y=\frac{\log x}{x} d x\) \(y d y=\log x d(\log x)\) On integrating, we get- \(\frac{y^{2}}{2}=\frac{(\log x)^{2}}{2}+c\) \(y^{2}=\left(\log _{e} x\right)^{2}+2 c\) \(y= \pm \sqrt{\left(\log _{e} x\right)^{2}+2 c}\)
Manipal UGET-2010
Differential Equation
87563
The solution of the differential equation \(\frac{d y}{d x}=\sin (x+y) \tan (x+y)-1\) is
1 \(\operatorname{cosec}(x+y)+\tan (x+y)=x+C\)
2 \(x+\operatorname{cosec}(x+y)=C\)
3 \(x+\tan (x+y)=C\)
4 \(x+\sec (x+y)=C\)
Explanation:
(B) : We have, \(\frac{d y}{d x}=\sin (x+y) \tan (x+y)-1\) Put, \(\quad x+y=v\), we get- \(1+\frac{d y}{d x}=\frac{d v}{d x}\) \(\frac{d y}{d x}=\frac{d v}{d x}-1\) Equation (i) reducing to, \(\frac{\mathrm{dv}}{\mathrm{dx}}=\sin \mathrm{v} \tan \mathrm{v}\) \(\frac{d v}{\sin v \tan v}=d x\) \(\int \operatorname{cosec} v \cot \mathrm{v} d \mathrm{v}=\int \mathrm{dx}\) \(-\operatorname{cosec} v \operatorname{cotv}=\mathrm{x}+\mathrm{C}\) \(-\operatorname{cosec}(\mathrm{x}+\mathrm{y})=\mathrm{x}+\mathrm{C}\) \(\mathrm{x}+\operatorname{cosec}(\mathrm{x}+\mathrm{y})=\mathrm{C}\)
Manipal UGET-2016
Differential Equation
87564
The solution of differential equation \((y \log x-1) y d x=x d y\) is
1 \(y\left(\log e^{x}+C x\right)=1\)
2 \(\left(\log \frac{x}{e}+C x\right) x=y\)
3 \(\left(\log C x^{2}+e x^{2}\right) y=x\)
4 None of these
Explanation:
(D) : Given differential equation is:- \((y \log x-1) y d x=x d y\) \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{(\mathrm{y} \log \mathrm{x}-1) \mathrm{y}}{\mathrm{x}}\) \(\frac{d y}{d x}=\frac{y^{2} \log x}{x}-\frac{y}{x}\) \(\frac{d y}{d x}+\frac{y}{x}=\frac{y^{2} \log x}{x}\) \(\frac{1}{y^2} \frac{d y}{d x}+\frac{y^{-1}}{x}=\frac{\log x}{x}\) Put, \(y^{-1}=v\) \(-y^{-2} \frac{d y}{d x}=\frac{d v}{d x}\) \(y^{-2} \frac{d y}{d x}=-\frac{d v}{d x}\) From \(\mathrm{Eq}^{\mathrm{n}}\) (i), we have:- \(-\frac{d v}{d x}+\frac{v}{x}=\frac{\log x}{x} \Rightarrow \frac{d y}{d x}-\frac{v}{x}=-\frac{\log x}{x}\) Here \(P=\frac{-1}{x}, Q=-\frac{\log x}{x}\) \(I \cdot F \cdot=e^{\int-\frac{1}{x} d x}=e^{-\log x}=\frac{1}{x}\) So, \(\mathrm{v} \cdot \frac{1}{\mathrm{x}}=\int \frac{1}{\mathrm{x}}\left(-\frac{\log \mathrm{x}}{\mathrm{x}}\right) \mathrm{dx}+\mathrm{C}\) \(=-\left[\log \mathrm{x}\left(\frac{-1}{\mathrm{x}}\right)+\int \frac{1}{\mathrm{x}} \cdot \frac{1}{\mathrm{x}} \mathrm{dx}\right]+C\) \(\frac{1}{\mathrm{x} \cdot \mathrm{y}}=\frac{\log \mathrm{x}}{\mathrm{x}}+\frac{1}{\mathrm{x}}+\mathrm{C} \quad\left[\because \mathrm{v}=\frac{1}{\mathrm{y}}\right]\) \(1=\mathrm{y}[\log \mathrm{x}+1+\mathrm{Cx}]\) \(1=\mathrm{y}[\log \mathrm{x}+\log \mathrm{e}+\mathrm{Cx}]\) \(1=y[\log (\mathrm{e} \cdot \mathrm{x})+\mathrm{cx}]\)
Manipal UGET-2020
Differential Equation
87565
Solution of the equation \(\cos ^{2} x \frac{d y}{d x}-(\tan 2 x) y=\cos ^{4} x,|x|\lt \frac{\pi}{4}\), where \(y\left(\frac{\pi}{6}\right)=\frac{3 \sqrt{3}}{8}\), is given by
(C) : The given differential equation can be written as- \(x=e^{x y \frac{d y}{d x}}\) \(\log x=x y \frac{d y}{d x}\) \(y d y=\frac{\log x}{x} d x\) \(y d y=\log x d(\log x)\) On integrating, we get- \(\frac{y^{2}}{2}=\frac{(\log x)^{2}}{2}+c\) \(y^{2}=\left(\log _{e} x\right)^{2}+2 c\) \(y= \pm \sqrt{\left(\log _{e} x\right)^{2}+2 c}\)
Manipal UGET-2010
Differential Equation
87563
The solution of the differential equation \(\frac{d y}{d x}=\sin (x+y) \tan (x+y)-1\) is
1 \(\operatorname{cosec}(x+y)+\tan (x+y)=x+C\)
2 \(x+\operatorname{cosec}(x+y)=C\)
3 \(x+\tan (x+y)=C\)
4 \(x+\sec (x+y)=C\)
Explanation:
(B) : We have, \(\frac{d y}{d x}=\sin (x+y) \tan (x+y)-1\) Put, \(\quad x+y=v\), we get- \(1+\frac{d y}{d x}=\frac{d v}{d x}\) \(\frac{d y}{d x}=\frac{d v}{d x}-1\) Equation (i) reducing to, \(\frac{\mathrm{dv}}{\mathrm{dx}}=\sin \mathrm{v} \tan \mathrm{v}\) \(\frac{d v}{\sin v \tan v}=d x\) \(\int \operatorname{cosec} v \cot \mathrm{v} d \mathrm{v}=\int \mathrm{dx}\) \(-\operatorname{cosec} v \operatorname{cotv}=\mathrm{x}+\mathrm{C}\) \(-\operatorname{cosec}(\mathrm{x}+\mathrm{y})=\mathrm{x}+\mathrm{C}\) \(\mathrm{x}+\operatorname{cosec}(\mathrm{x}+\mathrm{y})=\mathrm{C}\)
Manipal UGET-2016
Differential Equation
87564
The solution of differential equation \((y \log x-1) y d x=x d y\) is
1 \(y\left(\log e^{x}+C x\right)=1\)
2 \(\left(\log \frac{x}{e}+C x\right) x=y\)
3 \(\left(\log C x^{2}+e x^{2}\right) y=x\)
4 None of these
Explanation:
(D) : Given differential equation is:- \((y \log x-1) y d x=x d y\) \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{(\mathrm{y} \log \mathrm{x}-1) \mathrm{y}}{\mathrm{x}}\) \(\frac{d y}{d x}=\frac{y^{2} \log x}{x}-\frac{y}{x}\) \(\frac{d y}{d x}+\frac{y}{x}=\frac{y^{2} \log x}{x}\) \(\frac{1}{y^2} \frac{d y}{d x}+\frac{y^{-1}}{x}=\frac{\log x}{x}\) Put, \(y^{-1}=v\) \(-y^{-2} \frac{d y}{d x}=\frac{d v}{d x}\) \(y^{-2} \frac{d y}{d x}=-\frac{d v}{d x}\) From \(\mathrm{Eq}^{\mathrm{n}}\) (i), we have:- \(-\frac{d v}{d x}+\frac{v}{x}=\frac{\log x}{x} \Rightarrow \frac{d y}{d x}-\frac{v}{x}=-\frac{\log x}{x}\) Here \(P=\frac{-1}{x}, Q=-\frac{\log x}{x}\) \(I \cdot F \cdot=e^{\int-\frac{1}{x} d x}=e^{-\log x}=\frac{1}{x}\) So, \(\mathrm{v} \cdot \frac{1}{\mathrm{x}}=\int \frac{1}{\mathrm{x}}\left(-\frac{\log \mathrm{x}}{\mathrm{x}}\right) \mathrm{dx}+\mathrm{C}\) \(=-\left[\log \mathrm{x}\left(\frac{-1}{\mathrm{x}}\right)+\int \frac{1}{\mathrm{x}} \cdot \frac{1}{\mathrm{x}} \mathrm{dx}\right]+C\) \(\frac{1}{\mathrm{x} \cdot \mathrm{y}}=\frac{\log \mathrm{x}}{\mathrm{x}}+\frac{1}{\mathrm{x}}+\mathrm{C} \quad\left[\because \mathrm{v}=\frac{1}{\mathrm{y}}\right]\) \(1=\mathrm{y}[\log \mathrm{x}+1+\mathrm{Cx}]\) \(1=\mathrm{y}[\log \mathrm{x}+\log \mathrm{e}+\mathrm{Cx}]\) \(1=y[\log (\mathrm{e} \cdot \mathrm{x})+\mathrm{cx}]\)
Manipal UGET-2020
Differential Equation
87565
Solution of the equation \(\cos ^{2} x \frac{d y}{d x}-(\tan 2 x) y=\cos ^{4} x,|x|\lt \frac{\pi}{4}\), where \(y\left(\frac{\pi}{6}\right)=\frac{3 \sqrt{3}}{8}\), is given by
(C) : The given differential equation can be written as- \(x=e^{x y \frac{d y}{d x}}\) \(\log x=x y \frac{d y}{d x}\) \(y d y=\frac{\log x}{x} d x\) \(y d y=\log x d(\log x)\) On integrating, we get- \(\frac{y^{2}}{2}=\frac{(\log x)^{2}}{2}+c\) \(y^{2}=\left(\log _{e} x\right)^{2}+2 c\) \(y= \pm \sqrt{\left(\log _{e} x\right)^{2}+2 c}\)
Manipal UGET-2010
Differential Equation
87563
The solution of the differential equation \(\frac{d y}{d x}=\sin (x+y) \tan (x+y)-1\) is
1 \(\operatorname{cosec}(x+y)+\tan (x+y)=x+C\)
2 \(x+\operatorname{cosec}(x+y)=C\)
3 \(x+\tan (x+y)=C\)
4 \(x+\sec (x+y)=C\)
Explanation:
(B) : We have, \(\frac{d y}{d x}=\sin (x+y) \tan (x+y)-1\) Put, \(\quad x+y=v\), we get- \(1+\frac{d y}{d x}=\frac{d v}{d x}\) \(\frac{d y}{d x}=\frac{d v}{d x}-1\) Equation (i) reducing to, \(\frac{\mathrm{dv}}{\mathrm{dx}}=\sin \mathrm{v} \tan \mathrm{v}\) \(\frac{d v}{\sin v \tan v}=d x\) \(\int \operatorname{cosec} v \cot \mathrm{v} d \mathrm{v}=\int \mathrm{dx}\) \(-\operatorname{cosec} v \operatorname{cotv}=\mathrm{x}+\mathrm{C}\) \(-\operatorname{cosec}(\mathrm{x}+\mathrm{y})=\mathrm{x}+\mathrm{C}\) \(\mathrm{x}+\operatorname{cosec}(\mathrm{x}+\mathrm{y})=\mathrm{C}\)
Manipal UGET-2016
Differential Equation
87564
The solution of differential equation \((y \log x-1) y d x=x d y\) is
1 \(y\left(\log e^{x}+C x\right)=1\)
2 \(\left(\log \frac{x}{e}+C x\right) x=y\)
3 \(\left(\log C x^{2}+e x^{2}\right) y=x\)
4 None of these
Explanation:
(D) : Given differential equation is:- \((y \log x-1) y d x=x d y\) \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{(\mathrm{y} \log \mathrm{x}-1) \mathrm{y}}{\mathrm{x}}\) \(\frac{d y}{d x}=\frac{y^{2} \log x}{x}-\frac{y}{x}\) \(\frac{d y}{d x}+\frac{y}{x}=\frac{y^{2} \log x}{x}\) \(\frac{1}{y^2} \frac{d y}{d x}+\frac{y^{-1}}{x}=\frac{\log x}{x}\) Put, \(y^{-1}=v\) \(-y^{-2} \frac{d y}{d x}=\frac{d v}{d x}\) \(y^{-2} \frac{d y}{d x}=-\frac{d v}{d x}\) From \(\mathrm{Eq}^{\mathrm{n}}\) (i), we have:- \(-\frac{d v}{d x}+\frac{v}{x}=\frac{\log x}{x} \Rightarrow \frac{d y}{d x}-\frac{v}{x}=-\frac{\log x}{x}\) Here \(P=\frac{-1}{x}, Q=-\frac{\log x}{x}\) \(I \cdot F \cdot=e^{\int-\frac{1}{x} d x}=e^{-\log x}=\frac{1}{x}\) So, \(\mathrm{v} \cdot \frac{1}{\mathrm{x}}=\int \frac{1}{\mathrm{x}}\left(-\frac{\log \mathrm{x}}{\mathrm{x}}\right) \mathrm{dx}+\mathrm{C}\) \(=-\left[\log \mathrm{x}\left(\frac{-1}{\mathrm{x}}\right)+\int \frac{1}{\mathrm{x}} \cdot \frac{1}{\mathrm{x}} \mathrm{dx}\right]+C\) \(\frac{1}{\mathrm{x} \cdot \mathrm{y}}=\frac{\log \mathrm{x}}{\mathrm{x}}+\frac{1}{\mathrm{x}}+\mathrm{C} \quad\left[\because \mathrm{v}=\frac{1}{\mathrm{y}}\right]\) \(1=\mathrm{y}[\log \mathrm{x}+1+\mathrm{Cx}]\) \(1=\mathrm{y}[\log \mathrm{x}+\log \mathrm{e}+\mathrm{Cx}]\) \(1=y[\log (\mathrm{e} \cdot \mathrm{x})+\mathrm{cx}]\)
Manipal UGET-2020
Differential Equation
87565
Solution of the equation \(\cos ^{2} x \frac{d y}{d x}-(\tan 2 x) y=\cos ^{4} x,|x|\lt \frac{\pi}{4}\), where \(y\left(\frac{\pi}{6}\right)=\frac{3 \sqrt{3}}{8}\), is given by
(C) : The given differential equation can be written as- \(x=e^{x y \frac{d y}{d x}}\) \(\log x=x y \frac{d y}{d x}\) \(y d y=\frac{\log x}{x} d x\) \(y d y=\log x d(\log x)\) On integrating, we get- \(\frac{y^{2}}{2}=\frac{(\log x)^{2}}{2}+c\) \(y^{2}=\left(\log _{e} x\right)^{2}+2 c\) \(y= \pm \sqrt{\left(\log _{e} x\right)^{2}+2 c}\)
Manipal UGET-2010
Differential Equation
87563
The solution of the differential equation \(\frac{d y}{d x}=\sin (x+y) \tan (x+y)-1\) is
1 \(\operatorname{cosec}(x+y)+\tan (x+y)=x+C\)
2 \(x+\operatorname{cosec}(x+y)=C\)
3 \(x+\tan (x+y)=C\)
4 \(x+\sec (x+y)=C\)
Explanation:
(B) : We have, \(\frac{d y}{d x}=\sin (x+y) \tan (x+y)-1\) Put, \(\quad x+y=v\), we get- \(1+\frac{d y}{d x}=\frac{d v}{d x}\) \(\frac{d y}{d x}=\frac{d v}{d x}-1\) Equation (i) reducing to, \(\frac{\mathrm{dv}}{\mathrm{dx}}=\sin \mathrm{v} \tan \mathrm{v}\) \(\frac{d v}{\sin v \tan v}=d x\) \(\int \operatorname{cosec} v \cot \mathrm{v} d \mathrm{v}=\int \mathrm{dx}\) \(-\operatorname{cosec} v \operatorname{cotv}=\mathrm{x}+\mathrm{C}\) \(-\operatorname{cosec}(\mathrm{x}+\mathrm{y})=\mathrm{x}+\mathrm{C}\) \(\mathrm{x}+\operatorname{cosec}(\mathrm{x}+\mathrm{y})=\mathrm{C}\)
Manipal UGET-2016
Differential Equation
87564
The solution of differential equation \((y \log x-1) y d x=x d y\) is
1 \(y\left(\log e^{x}+C x\right)=1\)
2 \(\left(\log \frac{x}{e}+C x\right) x=y\)
3 \(\left(\log C x^{2}+e x^{2}\right) y=x\)
4 None of these
Explanation:
(D) : Given differential equation is:- \((y \log x-1) y d x=x d y\) \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{(\mathrm{y} \log \mathrm{x}-1) \mathrm{y}}{\mathrm{x}}\) \(\frac{d y}{d x}=\frac{y^{2} \log x}{x}-\frac{y}{x}\) \(\frac{d y}{d x}+\frac{y}{x}=\frac{y^{2} \log x}{x}\) \(\frac{1}{y^2} \frac{d y}{d x}+\frac{y^{-1}}{x}=\frac{\log x}{x}\) Put, \(y^{-1}=v\) \(-y^{-2} \frac{d y}{d x}=\frac{d v}{d x}\) \(y^{-2} \frac{d y}{d x}=-\frac{d v}{d x}\) From \(\mathrm{Eq}^{\mathrm{n}}\) (i), we have:- \(-\frac{d v}{d x}+\frac{v}{x}=\frac{\log x}{x} \Rightarrow \frac{d y}{d x}-\frac{v}{x}=-\frac{\log x}{x}\) Here \(P=\frac{-1}{x}, Q=-\frac{\log x}{x}\) \(I \cdot F \cdot=e^{\int-\frac{1}{x} d x}=e^{-\log x}=\frac{1}{x}\) So, \(\mathrm{v} \cdot \frac{1}{\mathrm{x}}=\int \frac{1}{\mathrm{x}}\left(-\frac{\log \mathrm{x}}{\mathrm{x}}\right) \mathrm{dx}+\mathrm{C}\) \(=-\left[\log \mathrm{x}\left(\frac{-1}{\mathrm{x}}\right)+\int \frac{1}{\mathrm{x}} \cdot \frac{1}{\mathrm{x}} \mathrm{dx}\right]+C\) \(\frac{1}{\mathrm{x} \cdot \mathrm{y}}=\frac{\log \mathrm{x}}{\mathrm{x}}+\frac{1}{\mathrm{x}}+\mathrm{C} \quad\left[\because \mathrm{v}=\frac{1}{\mathrm{y}}\right]\) \(1=\mathrm{y}[\log \mathrm{x}+1+\mathrm{Cx}]\) \(1=\mathrm{y}[\log \mathrm{x}+\log \mathrm{e}+\mathrm{Cx}]\) \(1=y[\log (\mathrm{e} \cdot \mathrm{x})+\mathrm{cx}]\)
Manipal UGET-2020
Differential Equation
87565
Solution of the equation \(\cos ^{2} x \frac{d y}{d x}-(\tan 2 x) y=\cos ^{4} x,|x|\lt \frac{\pi}{4}\), where \(y\left(\frac{\pi}{6}\right)=\frac{3 \sqrt{3}}{8}\), is given by
