(A) : Given equation is \(\left(x^{2}+y^{2}\right) d x=2 x y d y\) \(\frac{d y}{d x}=\frac{x^{2}+y^{2}}{2 x y}\) Which is homogeneous differential equation, Let, \(\quad y=v x\) \(\frac{d y}{d x}=v+x \frac{d v}{d x}\) \(v+x \frac{d v}{d x}=\frac{v^{2} x^{2}+x^{2}}{(2 x)(v x)}\) \(\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{\mathrm{v}^{2}+1}{2 \mathrm{v}}-\mathrm{v}\) \(\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{\mathrm{v}^{2}+1-2 \mathrm{v}^{2}}{2 \mathrm{v}}\) \(\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{1-\mathrm{v}^{2}}{2 \mathrm{v}}\) \(\int-\frac{2 v}{1-v^{2}} d v=\int-\frac{d x}{x}\) On integration both side, we get- \(-\log \left(1-v^{2}\right)=-(\log x+\log c)\) \(\log \left(1-v^{2}\right) =\log \left(\frac{c}{x}\right)\) \(\log \left(1-\frac{y^{2}}{x^{2}}\right) =\log \left(\frac{c}{x}\right)\) \(\frac{x^{2}-y^{2}}{x^{2}} =\frac{c}{x}\) We have, given that at \(\mathrm{x}=1, \mathrm{y}=0\) \(\therefore \mathrm{c}=1\) So, the required solution is, \(x^2-y^2=x\)
SRM JEEE-2013
Differential Equation
87504
In order to solve the differential equation \(x \cos x \frac{d y}{d x}+y(x \sin x+\cos x)=1\) The integrating factor is :
1 \(x \cos x\)
2 \(x \sec x\)
3 \(x \sin x\)
4 \(x \operatorname{cosec} x\)
Explanation:
(B) : Given differential equation is : \(x \cos x d y / d x+y(x \sin x+\cos x)=1\) Dividing both the sides by \(\mathrm{x} \cos \mathrm{x}\), \(\frac{d y}{d x}+\frac{x y \sin x}{x \cos x}+\frac{y \cos x}{x \cos x}=\frac{1}{x \cos x}\) \(\frac{d y}{d x}+y \tan x+\frac{y}{x}=\frac{1}{x \cos x}\) \(\frac{d y}{d x}+\left(\tan x+\frac{1}{x}\right) y=\frac{\sec x}{x}\) Which is of the form \(\frac{d y}{d x}+P y=Q\) Where, \(P=\tan x+\frac{1}{x}\) and \(Q=\frac{\sec x}{x}\) \(I F=e^{\int \operatorname{Pdx}}=\mathrm{e}^{\int \tan x+\frac{1}{x} d x}\) \(=\mathrm{e}^{(\log \sec \mathrm{x}+\log \mathrm{x})}=\mathrm{e}^{\log (\sec \mathrm{x} x)}=\mathrm{x} \sec \mathrm{x}\)
JCECE-2009
Differential Equation
87505
The solution of the differential equation \((x+1) \frac{d y}{d x}-y=e^{3 x}(x+1)^{2}\) is
(A) : Given equation is \(\left(x^{2}+y^{2}\right) d x=2 x y d y\) \(\frac{d y}{d x}=\frac{x^{2}+y^{2}}{2 x y}\) Which is homogeneous differential equation, Let, \(\quad y=v x\) \(\frac{d y}{d x}=v+x \frac{d v}{d x}\) \(v+x \frac{d v}{d x}=\frac{v^{2} x^{2}+x^{2}}{(2 x)(v x)}\) \(\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{\mathrm{v}^{2}+1}{2 \mathrm{v}}-\mathrm{v}\) \(\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{\mathrm{v}^{2}+1-2 \mathrm{v}^{2}}{2 \mathrm{v}}\) \(\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{1-\mathrm{v}^{2}}{2 \mathrm{v}}\) \(\int-\frac{2 v}{1-v^{2}} d v=\int-\frac{d x}{x}\) On integration both side, we get- \(-\log \left(1-v^{2}\right)=-(\log x+\log c)\) \(\log \left(1-v^{2}\right) =\log \left(\frac{c}{x}\right)\) \(\log \left(1-\frac{y^{2}}{x^{2}}\right) =\log \left(\frac{c}{x}\right)\) \(\frac{x^{2}-y^{2}}{x^{2}} =\frac{c}{x}\) We have, given that at \(\mathrm{x}=1, \mathrm{y}=0\) \(\therefore \mathrm{c}=1\) So, the required solution is, \(x^2-y^2=x\)
SRM JEEE-2013
Differential Equation
87504
In order to solve the differential equation \(x \cos x \frac{d y}{d x}+y(x \sin x+\cos x)=1\) The integrating factor is :
1 \(x \cos x\)
2 \(x \sec x\)
3 \(x \sin x\)
4 \(x \operatorname{cosec} x\)
Explanation:
(B) : Given differential equation is : \(x \cos x d y / d x+y(x \sin x+\cos x)=1\) Dividing both the sides by \(\mathrm{x} \cos \mathrm{x}\), \(\frac{d y}{d x}+\frac{x y \sin x}{x \cos x}+\frac{y \cos x}{x \cos x}=\frac{1}{x \cos x}\) \(\frac{d y}{d x}+y \tan x+\frac{y}{x}=\frac{1}{x \cos x}\) \(\frac{d y}{d x}+\left(\tan x+\frac{1}{x}\right) y=\frac{\sec x}{x}\) Which is of the form \(\frac{d y}{d x}+P y=Q\) Where, \(P=\tan x+\frac{1}{x}\) and \(Q=\frac{\sec x}{x}\) \(I F=e^{\int \operatorname{Pdx}}=\mathrm{e}^{\int \tan x+\frac{1}{x} d x}\) \(=\mathrm{e}^{(\log \sec \mathrm{x}+\log \mathrm{x})}=\mathrm{e}^{\log (\sec \mathrm{x} x)}=\mathrm{x} \sec \mathrm{x}\)
JCECE-2009
Differential Equation
87505
The solution of the differential equation \((x+1) \frac{d y}{d x}-y=e^{3 x}(x+1)^{2}\) is
(A) : Given equation is \(\left(x^{2}+y^{2}\right) d x=2 x y d y\) \(\frac{d y}{d x}=\frac{x^{2}+y^{2}}{2 x y}\) Which is homogeneous differential equation, Let, \(\quad y=v x\) \(\frac{d y}{d x}=v+x \frac{d v}{d x}\) \(v+x \frac{d v}{d x}=\frac{v^{2} x^{2}+x^{2}}{(2 x)(v x)}\) \(\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{\mathrm{v}^{2}+1}{2 \mathrm{v}}-\mathrm{v}\) \(\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{\mathrm{v}^{2}+1-2 \mathrm{v}^{2}}{2 \mathrm{v}}\) \(\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{1-\mathrm{v}^{2}}{2 \mathrm{v}}\) \(\int-\frac{2 v}{1-v^{2}} d v=\int-\frac{d x}{x}\) On integration both side, we get- \(-\log \left(1-v^{2}\right)=-(\log x+\log c)\) \(\log \left(1-v^{2}\right) =\log \left(\frac{c}{x}\right)\) \(\log \left(1-\frac{y^{2}}{x^{2}}\right) =\log \left(\frac{c}{x}\right)\) \(\frac{x^{2}-y^{2}}{x^{2}} =\frac{c}{x}\) We have, given that at \(\mathrm{x}=1, \mathrm{y}=0\) \(\therefore \mathrm{c}=1\) So, the required solution is, \(x^2-y^2=x\)
SRM JEEE-2013
Differential Equation
87504
In order to solve the differential equation \(x \cos x \frac{d y}{d x}+y(x \sin x+\cos x)=1\) The integrating factor is :
1 \(x \cos x\)
2 \(x \sec x\)
3 \(x \sin x\)
4 \(x \operatorname{cosec} x\)
Explanation:
(B) : Given differential equation is : \(x \cos x d y / d x+y(x \sin x+\cos x)=1\) Dividing both the sides by \(\mathrm{x} \cos \mathrm{x}\), \(\frac{d y}{d x}+\frac{x y \sin x}{x \cos x}+\frac{y \cos x}{x \cos x}=\frac{1}{x \cos x}\) \(\frac{d y}{d x}+y \tan x+\frac{y}{x}=\frac{1}{x \cos x}\) \(\frac{d y}{d x}+\left(\tan x+\frac{1}{x}\right) y=\frac{\sec x}{x}\) Which is of the form \(\frac{d y}{d x}+P y=Q\) Where, \(P=\tan x+\frac{1}{x}\) and \(Q=\frac{\sec x}{x}\) \(I F=e^{\int \operatorname{Pdx}}=\mathrm{e}^{\int \tan x+\frac{1}{x} d x}\) \(=\mathrm{e}^{(\log \sec \mathrm{x}+\log \mathrm{x})}=\mathrm{e}^{\log (\sec \mathrm{x} x)}=\mathrm{x} \sec \mathrm{x}\)
JCECE-2009
Differential Equation
87505
The solution of the differential equation \((x+1) \frac{d y}{d x}-y=e^{3 x}(x+1)^{2}\) is
(A) : Given equation is \(\left(x^{2}+y^{2}\right) d x=2 x y d y\) \(\frac{d y}{d x}=\frac{x^{2}+y^{2}}{2 x y}\) Which is homogeneous differential equation, Let, \(\quad y=v x\) \(\frac{d y}{d x}=v+x \frac{d v}{d x}\) \(v+x \frac{d v}{d x}=\frac{v^{2} x^{2}+x^{2}}{(2 x)(v x)}\) \(\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{\mathrm{v}^{2}+1}{2 \mathrm{v}}-\mathrm{v}\) \(\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{\mathrm{v}^{2}+1-2 \mathrm{v}^{2}}{2 \mathrm{v}}\) \(\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{1-\mathrm{v}^{2}}{2 \mathrm{v}}\) \(\int-\frac{2 v}{1-v^{2}} d v=\int-\frac{d x}{x}\) On integration both side, we get- \(-\log \left(1-v^{2}\right)=-(\log x+\log c)\) \(\log \left(1-v^{2}\right) =\log \left(\frac{c}{x}\right)\) \(\log \left(1-\frac{y^{2}}{x^{2}}\right) =\log \left(\frac{c}{x}\right)\) \(\frac{x^{2}-y^{2}}{x^{2}} =\frac{c}{x}\) We have, given that at \(\mathrm{x}=1, \mathrm{y}=0\) \(\therefore \mathrm{c}=1\) So, the required solution is, \(x^2-y^2=x\)
SRM JEEE-2013
Differential Equation
87504
In order to solve the differential equation \(x \cos x \frac{d y}{d x}+y(x \sin x+\cos x)=1\) The integrating factor is :
1 \(x \cos x\)
2 \(x \sec x\)
3 \(x \sin x\)
4 \(x \operatorname{cosec} x\)
Explanation:
(B) : Given differential equation is : \(x \cos x d y / d x+y(x \sin x+\cos x)=1\) Dividing both the sides by \(\mathrm{x} \cos \mathrm{x}\), \(\frac{d y}{d x}+\frac{x y \sin x}{x \cos x}+\frac{y \cos x}{x \cos x}=\frac{1}{x \cos x}\) \(\frac{d y}{d x}+y \tan x+\frac{y}{x}=\frac{1}{x \cos x}\) \(\frac{d y}{d x}+\left(\tan x+\frac{1}{x}\right) y=\frac{\sec x}{x}\) Which is of the form \(\frac{d y}{d x}+P y=Q\) Where, \(P=\tan x+\frac{1}{x}\) and \(Q=\frac{\sec x}{x}\) \(I F=e^{\int \operatorname{Pdx}}=\mathrm{e}^{\int \tan x+\frac{1}{x} d x}\) \(=\mathrm{e}^{(\log \sec \mathrm{x}+\log \mathrm{x})}=\mathrm{e}^{\log (\sec \mathrm{x} x)}=\mathrm{x} \sec \mathrm{x}\)
JCECE-2009
Differential Equation
87505
The solution of the differential equation \((x+1) \frac{d y}{d x}-y=e^{3 x}(x+1)^{2}\) is