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Differential Equations question

2020 · 6 Sep · Shift 1 · Q21
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  5. /2020 · 6 Sep · Shift 1 · Q21

Differential Equations question

2020 · 6 Sep · Shift 1 · Q21

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
The general solution of the differential equation 1+x2+y2+x2y2\sqrt {1 + {x^2} + {y^2} + {x^2}{y^2}}1+x2+y2+x2y2​ + xy dydx{{dy} \over {dx}}dxdy​ = 0 is : (where C is a constant of integration)
  1. A
    1+y2+1+x2=12log⁡e(1+x2−11+x2+1)+C\sqrt {1 + {y^2}} + \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} - 1} \over {\sqrt {1 + {x^2}} + 1}}} \right) + C1+y2​+1+x2​=21​loge​(1+x2​+11+x2​−1​)+C
  2. B
    1+y2−1+x2=12log⁡e(1+x2−11+x2+1)+C\sqrt {1 + {y^2}} - \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} - 1} \over {\sqrt {1 + {x^2}} + 1}}} \right) + C1+y2​−1+x2​=21​loge​(1+x2​+11+x2​−1​)+C
  3. C
    1+y2+1+x2=12log⁡e(1+x2+11+x2−1)+C\sqrt {1 + {y^2}} + \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} + 1} \over {\sqrt {1 + {x^2}} - 1}}} \right) + C1+y2​+1+x2​=21​loge​(1+x2​−11+x2​+1​)+C
  4. D
    1+y2−1+x2=12log⁡e(1+x2+11+x2−1)+C\sqrt {1 + {y^2}} - \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} + 1} \over {\sqrt {1 + {x^2}} - 1}}} \right) + C1+y2​−1+x2​=21​loge​(1+x2​−11+x2​+1​)+C
View written solutionFree

Correct answer: C

  1. Rewrite the differential equation

Given

1+x2+y2+x2y2+xydydx=0.\sqrt{1+x^2+y^2+x^2y^2}+xy\frac{dy}{dx}=0.1+x2+y2+x2y2​+xydxdy​=0.

Notice that

1+x2+y2+x2y2=(1+x2)(1+y2).1+x^2+y^2+x^2y^2=(1+x^2)(1+y^2).1+x2+y2+x2y2=(1+x2)(1+y2).

So the equation becomes

(1+x2)(1+y2)+xydydx=0.\sqrt{(1+x^2)(1+y^2)}+xy\frac{dy}{dx}=0.(1+x2)(1+y2)​+xydxdy​=0.

Hence,

1+x21+y2+xydydx=0.\sqrt{1+x^2}\sqrt{1+y^2}+xy\frac{dy}{dx}=0.1+x2​1+y2​+xydxdy​=0.
  1. Separate the variables

Move the derivative term:

xydydx=−1+x21+y2.xy\frac{dy}{dx}=-\sqrt{1+x^2}\sqrt{1+y^2}.xydxdy​=−1+x2​1+y2​.

Therefore,

dydx=−1+x2x⋅1+y2y.\frac{dy}{dx}=-\frac{\sqrt{1+x^2}}{x}\cdot \frac{\sqrt{1+y^2}}{y}.dxdy​=−x1+x2​​⋅y1+y2​​.

So,

y1+y2 dy=−1+x2x dx.\frac{y}{\sqrt{1+y^2}}\,dy=-\frac{\sqrt{1+x^2}}{x}\,dx.1+y2​y​dy=−x1+x2​​dx.

This is separable.

  1. Integrate both sides

Left side:

∫y1+y2 dy=1+y2.\int \frac{y}{\sqrt{1+y^2}}\,dy = \sqrt{1+y^2}.∫1+y2​y​dy=1+y2​.

So we need

∫1+x2x dx.\int \frac{\sqrt{1+x^2}}{x}\,dx.∫x1+x2​​dx.

Let

t=1+x2.t=\sqrt{1+x^2}.t=1+x2​.

Then

t2=1+x2  ⟹  2t dt=2x dx  ⟹  dx=tx dt.t^2=1+x^2 \implies 2t\,dt=2x\,dx \implies dx=\frac{t}{x}\,dt.t2=1+x2⟹2tdt=2xdx⟹dx=xt​dt.

Thus,

∫1+x2x dx=∫tx⋅txdt=∫t2x2dt.\int \frac{\sqrt{1+x^2}}{x}\,dx = \int \frac{t}{x}\cdot \frac{t}{x}dt =\int \frac{t^2}{x^2}dt.∫x1+x2​​dx=∫xt​⋅xt​dt=∫x2t2​dt.

Since

x2=t2−1,x^2=t^2-1,x2=t2−1,

we get

∫t2t2−1dt=∫(1+1t2−1)dt.\int \frac{t^2}{t^2-1}dt =\int \left(1+\frac{1}{t^2-1}\right)dt.∫t2−1t2​dt=∫(1+t2−11​)dt.

Now,

∫1t2−1dt=12ln⁡∣t−1t+1∣.\int \frac{1}{t^2-1}dt=\frac12\ln\left|\frac{t-1}{t+1}\right|.∫t2−11​dt=21​ln​t+1t−1​​.

Hence,

∫1+x2xdx=t+12ln⁡∣t−1t+1∣+C=1+x2+12ln⁡∣1+x2−11+x2+1∣+C.\int \frac{\sqrt{1+x^2}}{x}dx =t+\frac12\ln\left|\frac{t-1}{t+1}\right|+C =\sqrt{1+x^2}+\frac12\ln\left|\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}\right|+C.∫x1+x2​​dx=t+21​ln​t+1t−1​​+C=1+x2​+21​ln​1+x2​+11+x2​−1​​+C.

Therefore, from

∫y1+y2dy=−∫1+x2xdx,\int \frac{y}{\sqrt{1+y^2}}dy=-\int \frac{\sqrt{1+x^2}}{x}dx,∫1+y2​y​dy=−∫x1+x2​​dx,

we get

1+y2=−1+x2−12ln⁡∣1+x2−11+x2+1∣+C.\sqrt{1+y^2}=-\sqrt{1+x^2}-\frac12\ln\left|\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}\right|+C.1+y2​=−1+x2​−21​ln​1+x2​+11+x2​−1​​+C.

Rearranging,

1+y2+1+x2=−12ln⁡∣1+x2−11+x2+1∣+C.\sqrt{1+y^2}+\sqrt{1+x^2}=-\frac12\ln\left|\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}\right|+C.1+y2​+1+x2​=−21​ln​1+x2​+11+x2​−1​​+C.

Using

−ln⁡(ab)=ln⁡(ba),-\ln\left(\frac{a}{b}\right)=\ln\left(\frac{b}{a}\right),−ln(ba​)=ln(ab​),

we obtain

1+y2+1+x2=12ln⁡∣1+x2+11+x2−1∣+C.\sqrt{1+y^2}+\sqrt{1+x^2}=\frac12\ln\left|\frac{\sqrt{1+x^2}+1}{\sqrt{1+x^2}-1}\right|+C.1+y2​+1+x2​=21​ln​1+x2​−11+x2​+1​​+C.
  1. Match with the options

This matches Option C:

1+y2+1+x2=12log⁡e(1+x2+11+x2−1)+C.\sqrt {1 + {y^2}} + \sqrt {1 + {x^2}} = \frac {1}{2}\log_e\left( \frac{\sqrt {1 + {x^2}} + 1}{\sqrt {1 + {x^2}} - 1} \right) + C.1+y2​+1+x2​=21​loge​(1+x2​−11+x2​+1​)+C.
  1. Conclusion

The correct answer is:

C\boxed{\text{C}}C​
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