(A) : Given that, \(x^{2} \frac{d y}{d x}-x y=1+\cos y / x\) Let, \(\mathrm{y}=\mathrm{vx}\) \(\frac{d y}{d x}=v+x \frac{d v}{d x}\) Putting the value of \(\frac{d y}{d x}\) in equation (i), we get- \(x^{2}\left(v+x \frac{d v}{d x}\right)-v x^{2}=1+\cos v\) \(v x^{2}+x^{3} \frac{d v}{d x}-v x^{2}=1+\cos v\) \(x^{3} \frac{d v}{d x}=1+\cos v\) \(\frac{d v}{1+\cos v}=\frac{d x}{x^{3}}\) Integrating in both sides we get- \(\frac{1}{2} \int \sec ^{2} v / 2=\int \frac{1}{x^{3}} d x \Rightarrow \frac{1}{2} \tan \frac{v}{2} \times 2=\frac{x^{-2}}{-2}+C\) \(\frac{1}{2} \tan \frac{v}{2} \times 2=-\frac{1}{2 x^{2}}+C\) \(\text { Put } y=v x \text { or } v=y / x\) \(\tan y / 2 x=C-\frac{1}{2 x^{2}}\)
UPSEE-2013]**#
Differential Equation
87227
The solution of the differential equation \(x d y-y d x=\left(\sqrt{x^{2}+y^{2}}\right) d x\) is
1 \(y-\sqrt{x^{2}+y^{2}}=C x^{2}\)
2 \(y+\sqrt{x^{2}+y^{2}}=C x^{2}\)
3 \(y+\sqrt{x^{2}+y^{2}}+C x^{2}=0\)
4 None of these
Explanation:
(B) : Given, \(x d y-y d x=\left(\sqrt{x^{2}+y^{2}}\right) d x\) \(x d y=\left(y+\sqrt{x^{2}+y^{2}}\right) d x\) \(\frac{d y}{d x}=\frac{y+\sqrt{x^{2}+y^{2}}}{x} \tag{i}\) Now putting, \(\mathrm{y}=\mathrm{vx}\) \(\frac{d y}{d x}=v+x \frac{d v}{d x}\) Putting in equation (i), we get- \(v+x \frac{d v}{d x}=\frac{v x+\sqrt{x^{2}+v^{2} x^{2}}}{x}\) \(v+x \frac{d v}{d x}=v+\sqrt{1+v^{2}} \Rightarrow x \frac{d v}{d x}=\sqrt{1+v^{2}}\) \(\frac{d v}{\sqrt{1+v^{2}}}=\frac{d x}{x}\) Integrating both sides, we get - \(\int \frac{d v}{\sqrt{1+v^{2}}}=\int \frac{d x}{x}\) We know that, \(\log \left|\mathrm{v}+\sqrt{\mathrm{v}^{2}+1}\right|=\log \mathrm{x}+\log \mathrm{C}\) \(\log \left|\mathrm{v}+\sqrt{\mathrm{v}^{2}+1}\right|=\log \mathrm{xC}\) \(\mathrm{v}+\sqrt{\mathrm{v}^{2}+1}=\mathrm{xC}\) Putting, \(v=y / x\) we get - \(\frac{y}{x}+\sqrt{\frac{y^{2}}{x^{2}}+1}=x C\) \(y+\sqrt{x^{2}+y^{2}}=x^{2} C\)
VITEEE-2008
Differential Equation
87228
The solution of the equation \(\sin ^{-1}\left(\frac{d y}{d x}\right)=x+y\) is
1 \(\tan (x+y)+\sec (x+y)=x+C\)
2 \(\tan (x+y)-\sec (x+y)=x+C\)
3 \(\tan (x+y)+\sec (x+y)+x+C=0\)
4 None of the above
Explanation:
(B) : We have, \(\sin ^{-1}(d y / d x)=x+y\) \(\frac{d y}{d x}=\sin (x+y) \tag{i}\) Let, \(\quad \mathrm{x}+\mathrm{y}=\mathrm{v}\) \(1+\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dv}}{\mathrm{dx}}\) From equation (i), we get- \(\frac{\mathrm{dv}}{\mathrm{dx}}-1=\sin \mathrm{v}\) Separating the variable, we get- \(\frac{d v}{1+\sin v}=d x\) \(\frac{(1-\sin v)}{(1+\sin v)(1-\sin v)} d v=d x\) \(\frac{1-\sin v}{\cos ^{2} v} d v=d x\) \(\left(\frac{1}{\cos ^{2} v}-\frac{\sin v}{\cos ^{2} v}\right) d v=d x\) \(\left(\sec ^{2} v-\tan v \sec v\right) d v=d x\) Integrating both sides, we get\(\tan \mathrm{v}-\sec \mathrm{v}=\mathrm{x}+\mathrm{C}\) Putting \(\mathrm{v}=\mathrm{x}+\mathrm{y}\) we get - \(\tan (x+y)-\sec (x+y)=x+C\)
(A) : Given that, \(x^{2} \frac{d y}{d x}-x y=1+\cos y / x\) Let, \(\mathrm{y}=\mathrm{vx}\) \(\frac{d y}{d x}=v+x \frac{d v}{d x}\) Putting the value of \(\frac{d y}{d x}\) in equation (i), we get- \(x^{2}\left(v+x \frac{d v}{d x}\right)-v x^{2}=1+\cos v\) \(v x^{2}+x^{3} \frac{d v}{d x}-v x^{2}=1+\cos v\) \(x^{3} \frac{d v}{d x}=1+\cos v\) \(\frac{d v}{1+\cos v}=\frac{d x}{x^{3}}\) Integrating in both sides we get- \(\frac{1}{2} \int \sec ^{2} v / 2=\int \frac{1}{x^{3}} d x \Rightarrow \frac{1}{2} \tan \frac{v}{2} \times 2=\frac{x^{-2}}{-2}+C\) \(\frac{1}{2} \tan \frac{v}{2} \times 2=-\frac{1}{2 x^{2}}+C\) \(\text { Put } y=v x \text { or } v=y / x\) \(\tan y / 2 x=C-\frac{1}{2 x^{2}}\)
UPSEE-2013]**#
Differential Equation
87227
The solution of the differential equation \(x d y-y d x=\left(\sqrt{x^{2}+y^{2}}\right) d x\) is
1 \(y-\sqrt{x^{2}+y^{2}}=C x^{2}\)
2 \(y+\sqrt{x^{2}+y^{2}}=C x^{2}\)
3 \(y+\sqrt{x^{2}+y^{2}}+C x^{2}=0\)
4 None of these
Explanation:
(B) : Given, \(x d y-y d x=\left(\sqrt{x^{2}+y^{2}}\right) d x\) \(x d y=\left(y+\sqrt{x^{2}+y^{2}}\right) d x\) \(\frac{d y}{d x}=\frac{y+\sqrt{x^{2}+y^{2}}}{x} \tag{i}\) Now putting, \(\mathrm{y}=\mathrm{vx}\) \(\frac{d y}{d x}=v+x \frac{d v}{d x}\) Putting in equation (i), we get- \(v+x \frac{d v}{d x}=\frac{v x+\sqrt{x^{2}+v^{2} x^{2}}}{x}\) \(v+x \frac{d v}{d x}=v+\sqrt{1+v^{2}} \Rightarrow x \frac{d v}{d x}=\sqrt{1+v^{2}}\) \(\frac{d v}{\sqrt{1+v^{2}}}=\frac{d x}{x}\) Integrating both sides, we get - \(\int \frac{d v}{\sqrt{1+v^{2}}}=\int \frac{d x}{x}\) We know that, \(\log \left|\mathrm{v}+\sqrt{\mathrm{v}^{2}+1}\right|=\log \mathrm{x}+\log \mathrm{C}\) \(\log \left|\mathrm{v}+\sqrt{\mathrm{v}^{2}+1}\right|=\log \mathrm{xC}\) \(\mathrm{v}+\sqrt{\mathrm{v}^{2}+1}=\mathrm{xC}\) Putting, \(v=y / x\) we get - \(\frac{y}{x}+\sqrt{\frac{y^{2}}{x^{2}}+1}=x C\) \(y+\sqrt{x^{2}+y^{2}}=x^{2} C\)
VITEEE-2008
Differential Equation
87228
The solution of the equation \(\sin ^{-1}\left(\frac{d y}{d x}\right)=x+y\) is
1 \(\tan (x+y)+\sec (x+y)=x+C\)
2 \(\tan (x+y)-\sec (x+y)=x+C\)
3 \(\tan (x+y)+\sec (x+y)+x+C=0\)
4 None of the above
Explanation:
(B) : We have, \(\sin ^{-1}(d y / d x)=x+y\) \(\frac{d y}{d x}=\sin (x+y) \tag{i}\) Let, \(\quad \mathrm{x}+\mathrm{y}=\mathrm{v}\) \(1+\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dv}}{\mathrm{dx}}\) From equation (i), we get- \(\frac{\mathrm{dv}}{\mathrm{dx}}-1=\sin \mathrm{v}\) Separating the variable, we get- \(\frac{d v}{1+\sin v}=d x\) \(\frac{(1-\sin v)}{(1+\sin v)(1-\sin v)} d v=d x\) \(\frac{1-\sin v}{\cos ^{2} v} d v=d x\) \(\left(\frac{1}{\cos ^{2} v}-\frac{\sin v}{\cos ^{2} v}\right) d v=d x\) \(\left(\sec ^{2} v-\tan v \sec v\right) d v=d x\) Integrating both sides, we get\(\tan \mathrm{v}-\sec \mathrm{v}=\mathrm{x}+\mathrm{C}\) Putting \(\mathrm{v}=\mathrm{x}+\mathrm{y}\) we get - \(\tan (x+y)-\sec (x+y)=x+C\)
(A) : Given that, \(x^{2} \frac{d y}{d x}-x y=1+\cos y / x\) Let, \(\mathrm{y}=\mathrm{vx}\) \(\frac{d y}{d x}=v+x \frac{d v}{d x}\) Putting the value of \(\frac{d y}{d x}\) in equation (i), we get- \(x^{2}\left(v+x \frac{d v}{d x}\right)-v x^{2}=1+\cos v\) \(v x^{2}+x^{3} \frac{d v}{d x}-v x^{2}=1+\cos v\) \(x^{3} \frac{d v}{d x}=1+\cos v\) \(\frac{d v}{1+\cos v}=\frac{d x}{x^{3}}\) Integrating in both sides we get- \(\frac{1}{2} \int \sec ^{2} v / 2=\int \frac{1}{x^{3}} d x \Rightarrow \frac{1}{2} \tan \frac{v}{2} \times 2=\frac{x^{-2}}{-2}+C\) \(\frac{1}{2} \tan \frac{v}{2} \times 2=-\frac{1}{2 x^{2}}+C\) \(\text { Put } y=v x \text { or } v=y / x\) \(\tan y / 2 x=C-\frac{1}{2 x^{2}}\)
UPSEE-2013]**#
Differential Equation
87227
The solution of the differential equation \(x d y-y d x=\left(\sqrt{x^{2}+y^{2}}\right) d x\) is
1 \(y-\sqrt{x^{2}+y^{2}}=C x^{2}\)
2 \(y+\sqrt{x^{2}+y^{2}}=C x^{2}\)
3 \(y+\sqrt{x^{2}+y^{2}}+C x^{2}=0\)
4 None of these
Explanation:
(B) : Given, \(x d y-y d x=\left(\sqrt{x^{2}+y^{2}}\right) d x\) \(x d y=\left(y+\sqrt{x^{2}+y^{2}}\right) d x\) \(\frac{d y}{d x}=\frac{y+\sqrt{x^{2}+y^{2}}}{x} \tag{i}\) Now putting, \(\mathrm{y}=\mathrm{vx}\) \(\frac{d y}{d x}=v+x \frac{d v}{d x}\) Putting in equation (i), we get- \(v+x \frac{d v}{d x}=\frac{v x+\sqrt{x^{2}+v^{2} x^{2}}}{x}\) \(v+x \frac{d v}{d x}=v+\sqrt{1+v^{2}} \Rightarrow x \frac{d v}{d x}=\sqrt{1+v^{2}}\) \(\frac{d v}{\sqrt{1+v^{2}}}=\frac{d x}{x}\) Integrating both sides, we get - \(\int \frac{d v}{\sqrt{1+v^{2}}}=\int \frac{d x}{x}\) We know that, \(\log \left|\mathrm{v}+\sqrt{\mathrm{v}^{2}+1}\right|=\log \mathrm{x}+\log \mathrm{C}\) \(\log \left|\mathrm{v}+\sqrt{\mathrm{v}^{2}+1}\right|=\log \mathrm{xC}\) \(\mathrm{v}+\sqrt{\mathrm{v}^{2}+1}=\mathrm{xC}\) Putting, \(v=y / x\) we get - \(\frac{y}{x}+\sqrt{\frac{y^{2}}{x^{2}}+1}=x C\) \(y+\sqrt{x^{2}+y^{2}}=x^{2} C\)
VITEEE-2008
Differential Equation
87228
The solution of the equation \(\sin ^{-1}\left(\frac{d y}{d x}\right)=x+y\) is
1 \(\tan (x+y)+\sec (x+y)=x+C\)
2 \(\tan (x+y)-\sec (x+y)=x+C\)
3 \(\tan (x+y)+\sec (x+y)+x+C=0\)
4 None of the above
Explanation:
(B) : We have, \(\sin ^{-1}(d y / d x)=x+y\) \(\frac{d y}{d x}=\sin (x+y) \tag{i}\) Let, \(\quad \mathrm{x}+\mathrm{y}=\mathrm{v}\) \(1+\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dv}}{\mathrm{dx}}\) From equation (i), we get- \(\frac{\mathrm{dv}}{\mathrm{dx}}-1=\sin \mathrm{v}\) Separating the variable, we get- \(\frac{d v}{1+\sin v}=d x\) \(\frac{(1-\sin v)}{(1+\sin v)(1-\sin v)} d v=d x\) \(\frac{1-\sin v}{\cos ^{2} v} d v=d x\) \(\left(\frac{1}{\cos ^{2} v}-\frac{\sin v}{\cos ^{2} v}\right) d v=d x\) \(\left(\sec ^{2} v-\tan v \sec v\right) d v=d x\) Integrating both sides, we get\(\tan \mathrm{v}-\sec \mathrm{v}=\mathrm{x}+\mathrm{C}\) Putting \(\mathrm{v}=\mathrm{x}+\mathrm{y}\) we get - \(\tan (x+y)-\sec (x+y)=x+C\)
(A) : Given that, \(x^{2} \frac{d y}{d x}-x y=1+\cos y / x\) Let, \(\mathrm{y}=\mathrm{vx}\) \(\frac{d y}{d x}=v+x \frac{d v}{d x}\) Putting the value of \(\frac{d y}{d x}\) in equation (i), we get- \(x^{2}\left(v+x \frac{d v}{d x}\right)-v x^{2}=1+\cos v\) \(v x^{2}+x^{3} \frac{d v}{d x}-v x^{2}=1+\cos v\) \(x^{3} \frac{d v}{d x}=1+\cos v\) \(\frac{d v}{1+\cos v}=\frac{d x}{x^{3}}\) Integrating in both sides we get- \(\frac{1}{2} \int \sec ^{2} v / 2=\int \frac{1}{x^{3}} d x \Rightarrow \frac{1}{2} \tan \frac{v}{2} \times 2=\frac{x^{-2}}{-2}+C\) \(\frac{1}{2} \tan \frac{v}{2} \times 2=-\frac{1}{2 x^{2}}+C\) \(\text { Put } y=v x \text { or } v=y / x\) \(\tan y / 2 x=C-\frac{1}{2 x^{2}}\)
UPSEE-2013]**#
Differential Equation
87227
The solution of the differential equation \(x d y-y d x=\left(\sqrt{x^{2}+y^{2}}\right) d x\) is
1 \(y-\sqrt{x^{2}+y^{2}}=C x^{2}\)
2 \(y+\sqrt{x^{2}+y^{2}}=C x^{2}\)
3 \(y+\sqrt{x^{2}+y^{2}}+C x^{2}=0\)
4 None of these
Explanation:
(B) : Given, \(x d y-y d x=\left(\sqrt{x^{2}+y^{2}}\right) d x\) \(x d y=\left(y+\sqrt{x^{2}+y^{2}}\right) d x\) \(\frac{d y}{d x}=\frac{y+\sqrt{x^{2}+y^{2}}}{x} \tag{i}\) Now putting, \(\mathrm{y}=\mathrm{vx}\) \(\frac{d y}{d x}=v+x \frac{d v}{d x}\) Putting in equation (i), we get- \(v+x \frac{d v}{d x}=\frac{v x+\sqrt{x^{2}+v^{2} x^{2}}}{x}\) \(v+x \frac{d v}{d x}=v+\sqrt{1+v^{2}} \Rightarrow x \frac{d v}{d x}=\sqrt{1+v^{2}}\) \(\frac{d v}{\sqrt{1+v^{2}}}=\frac{d x}{x}\) Integrating both sides, we get - \(\int \frac{d v}{\sqrt{1+v^{2}}}=\int \frac{d x}{x}\) We know that, \(\log \left|\mathrm{v}+\sqrt{\mathrm{v}^{2}+1}\right|=\log \mathrm{x}+\log \mathrm{C}\) \(\log \left|\mathrm{v}+\sqrt{\mathrm{v}^{2}+1}\right|=\log \mathrm{xC}\) \(\mathrm{v}+\sqrt{\mathrm{v}^{2}+1}=\mathrm{xC}\) Putting, \(v=y / x\) we get - \(\frac{y}{x}+\sqrt{\frac{y^{2}}{x^{2}}+1}=x C\) \(y+\sqrt{x^{2}+y^{2}}=x^{2} C\)
VITEEE-2008
Differential Equation
87228
The solution of the equation \(\sin ^{-1}\left(\frac{d y}{d x}\right)=x+y\) is
1 \(\tan (x+y)+\sec (x+y)=x+C\)
2 \(\tan (x+y)-\sec (x+y)=x+C\)
3 \(\tan (x+y)+\sec (x+y)+x+C=0\)
4 None of the above
Explanation:
(B) : We have, \(\sin ^{-1}(d y / d x)=x+y\) \(\frac{d y}{d x}=\sin (x+y) \tag{i}\) Let, \(\quad \mathrm{x}+\mathrm{y}=\mathrm{v}\) \(1+\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dv}}{\mathrm{dx}}\) From equation (i), we get- \(\frac{\mathrm{dv}}{\mathrm{dx}}-1=\sin \mathrm{v}\) Separating the variable, we get- \(\frac{d v}{1+\sin v}=d x\) \(\frac{(1-\sin v)}{(1+\sin v)(1-\sin v)} d v=d x\) \(\frac{1-\sin v}{\cos ^{2} v} d v=d x\) \(\left(\frac{1}{\cos ^{2} v}-\frac{\sin v}{\cos ^{2} v}\right) d v=d x\) \(\left(\sec ^{2} v-\tan v \sec v\right) d v=d x\) Integrating both sides, we get\(\tan \mathrm{v}-\sec \mathrm{v}=\mathrm{x}+\mathrm{C}\) Putting \(\mathrm{v}=\mathrm{x}+\mathrm{y}\) we get - \(\tan (x+y)-\sec (x+y)=x+C\)
(A) : Given that, \(x^{2} \frac{d y}{d x}-x y=1+\cos y / x\) Let, \(\mathrm{y}=\mathrm{vx}\) \(\frac{d y}{d x}=v+x \frac{d v}{d x}\) Putting the value of \(\frac{d y}{d x}\) in equation (i), we get- \(x^{2}\left(v+x \frac{d v}{d x}\right)-v x^{2}=1+\cos v\) \(v x^{2}+x^{3} \frac{d v}{d x}-v x^{2}=1+\cos v\) \(x^{3} \frac{d v}{d x}=1+\cos v\) \(\frac{d v}{1+\cos v}=\frac{d x}{x^{3}}\) Integrating in both sides we get- \(\frac{1}{2} \int \sec ^{2} v / 2=\int \frac{1}{x^{3}} d x \Rightarrow \frac{1}{2} \tan \frac{v}{2} \times 2=\frac{x^{-2}}{-2}+C\) \(\frac{1}{2} \tan \frac{v}{2} \times 2=-\frac{1}{2 x^{2}}+C\) \(\text { Put } y=v x \text { or } v=y / x\) \(\tan y / 2 x=C-\frac{1}{2 x^{2}}\)
UPSEE-2013]**#
Differential Equation
87227
The solution of the differential equation \(x d y-y d x=\left(\sqrt{x^{2}+y^{2}}\right) d x\) is
1 \(y-\sqrt{x^{2}+y^{2}}=C x^{2}\)
2 \(y+\sqrt{x^{2}+y^{2}}=C x^{2}\)
3 \(y+\sqrt{x^{2}+y^{2}}+C x^{2}=0\)
4 None of these
Explanation:
(B) : Given, \(x d y-y d x=\left(\sqrt{x^{2}+y^{2}}\right) d x\) \(x d y=\left(y+\sqrt{x^{2}+y^{2}}\right) d x\) \(\frac{d y}{d x}=\frac{y+\sqrt{x^{2}+y^{2}}}{x} \tag{i}\) Now putting, \(\mathrm{y}=\mathrm{vx}\) \(\frac{d y}{d x}=v+x \frac{d v}{d x}\) Putting in equation (i), we get- \(v+x \frac{d v}{d x}=\frac{v x+\sqrt{x^{2}+v^{2} x^{2}}}{x}\) \(v+x \frac{d v}{d x}=v+\sqrt{1+v^{2}} \Rightarrow x \frac{d v}{d x}=\sqrt{1+v^{2}}\) \(\frac{d v}{\sqrt{1+v^{2}}}=\frac{d x}{x}\) Integrating both sides, we get - \(\int \frac{d v}{\sqrt{1+v^{2}}}=\int \frac{d x}{x}\) We know that, \(\log \left|\mathrm{v}+\sqrt{\mathrm{v}^{2}+1}\right|=\log \mathrm{x}+\log \mathrm{C}\) \(\log \left|\mathrm{v}+\sqrt{\mathrm{v}^{2}+1}\right|=\log \mathrm{xC}\) \(\mathrm{v}+\sqrt{\mathrm{v}^{2}+1}=\mathrm{xC}\) Putting, \(v=y / x\) we get - \(\frac{y}{x}+\sqrt{\frac{y^{2}}{x^{2}}+1}=x C\) \(y+\sqrt{x^{2}+y^{2}}=x^{2} C\)
VITEEE-2008
Differential Equation
87228
The solution of the equation \(\sin ^{-1}\left(\frac{d y}{d x}\right)=x+y\) is
1 \(\tan (x+y)+\sec (x+y)=x+C\)
2 \(\tan (x+y)-\sec (x+y)=x+C\)
3 \(\tan (x+y)+\sec (x+y)+x+C=0\)
4 None of the above
Explanation:
(B) : We have, \(\sin ^{-1}(d y / d x)=x+y\) \(\frac{d y}{d x}=\sin (x+y) \tag{i}\) Let, \(\quad \mathrm{x}+\mathrm{y}=\mathrm{v}\) \(1+\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dv}}{\mathrm{dx}}\) From equation (i), we get- \(\frac{\mathrm{dv}}{\mathrm{dx}}-1=\sin \mathrm{v}\) Separating the variable, we get- \(\frac{d v}{1+\sin v}=d x\) \(\frac{(1-\sin v)}{(1+\sin v)(1-\sin v)} d v=d x\) \(\frac{1-\sin v}{\cos ^{2} v} d v=d x\) \(\left(\frac{1}{\cos ^{2} v}-\frac{\sin v}{\cos ^{2} v}\right) d v=d x\) \(\left(\sec ^{2} v-\tan v \sec v\right) d v=d x\) Integrating both sides, we get\(\tan \mathrm{v}-\sec \mathrm{v}=\mathrm{x}+\mathrm{C}\) Putting \(\mathrm{v}=\mathrm{x}+\mathrm{y}\) we get - \(\tan (x+y)-\sec (x+y)=x+C\)