The general solution of the differential equation sqrt1+x^2+y^2+x^2y^2+xy(dy)/(dx)=0 is : (where C is a constant of integration)

sqrt1+y^2+sqrt1+x^2=(1)/(2)log_e (dfracsqrt1+x^2-1sqrt1+x^2+1 )+C

sqrt1+y^2-sqrt1+x^2=(1)/(2)log_e (dfracsqrt1+x^2+1sqrt1+x^2-1 )+C

sqrt1+y^2+sqrt1+x^2=(1)/(2)log_e (dfracsqrt1+x^2+1sqrt1+x^2-1 )+C

sqrt1+y^2-sqrt1+x^2=(1)/(2)log_e (dfracsqrt1+x^2-1sqrt1+x^2+1 )+C

Differential Equation: The general solution of the differential equation… | JEE Main 2020

JEE Main 2020 (Online) 6th September Morning ShiftQuestion 1

The general solution of the differential equation $\sqrt{1+x^{2}+y^{2}+x^{2}y^{2}}+xy\dfrac{dy}{dx}=0$ is : (where $C$ is a constant of integration)

Differential Equation

$\sqrt{1+y^{2}}+\sqrt{1+x^{2}}=\dfrac{1}{2}\log_e\left(\dfrac{\sqrt{1+x^{2}}-1}{\sqrt{1+x^{2}}+1}\right)+C$$\sqrt{1+y^{2}}-\sqrt{1+x^{2}}=\dfrac{1}{2}\log_e\left(\dfrac{\sqrt{1+x^{2}}+1}{\sqrt{1+x^{2}}-1}\right)+C$$\sqrt{1+y^{2}}+\sqrt{1+x^{2}}=\dfrac{1}{2}\log_e\left(\dfrac{\sqrt{1+x^{2}}+1}{\sqrt{1+x^{2}}-1}\right)+C$✓ Correct answer$\sqrt{1+y^{2}}-\sqrt{1+x^{2}}=\dfrac{1}{2}\log_e\left(\dfrac{\sqrt{1+x^{2}}-1}{\sqrt{1+x^{2}}+1}\right)+C$

Solution

Concept. $1+x^2+y^2+x^2y^2=(1+x^2)(1+y^2)$, so the variables separate. Formulas used. (1) $\int\dfrac{y\,dy}{\sqrt{1+y^2}}=\sqrt{1+y^2}$(2) With $u=\sqrt{1+x^2}$: $\int\dfrac{\sqrt{1+x^2}}{x}\,dx=\int\dfrac{u^2\,du}{u^2-1}=u+\dfrac12\ln\dfrac{u-1}{u+1}$Step 1 - Separate. $$\frac{y\,dy}{\sqrt{1+y^2}}=-\frac{\sqrt{1+x^2}}{x}\,dx.$$ Step 2 - Integrate. By (1) and (2): $$\sqrt{1+y^2}=-\sqrt{1+x^2}-\frac12\ln\frac{u-1}{u+1}+C.$$ Step 3 - Rearrange. Moving terms and flipping the log: $$\sqrt{1+y^2}+\sqrt{1+x^2}=\frac12\ln\frac{\sqrt{1+x^2}+1}{\sqrt{1+x^2}-1}+C.$$ Answer: (C)

Differential Equation - JEE Mathematics question (JEE Main 2020 (Online) 6th September Morning Shift)

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