Question

The general solution y(x) of the differential equation 

\(\rm {dy\over dx}= {x\sqrt{1-y^2}\over y\sqrt{1-x^2}}\)

1. y² - 1 = C ln (x² - 1)
2. y² - 1 = C(x² - 1)
3. y² + 1 = C(x² + 1)
4. ln (y² - 1) = C(x² - 1)

Answer 1

Correct Answer - Option 2 : y² - 1 = C(x² - 1)

Concept:

Separation of variable:

If the equation is such that the variables can be separated, then

• Separate the variables.
• Take a single variable on either side.
• Integrate both sides w.r.t the respective variable.

Calculation:

The given differential equation is 

\(\rm {dy\over dx}= {x\sqrt{1-y^2}\over y\sqrt{1-x^2}}\)

\(\rm {ydy\over \sqrt{1-y^2}}= {xdx\over\sqrt{1-x^2}}\)

Integrating both the sides 

\(\rm \int{ydy\over \sqrt{1-y^2}}= \int{xdx\over\sqrt{1-x^2}}\)

\(\rm {-\ln\sqrt{1-y^2}}= -{\ln\sqrt{1-x^2}}+c\)

\(\rm {\ln\sqrt{1-y^2}} -{\ln\sqrt{1-x^2}}+c = 0\)

\(\rm \ln{\sqrt{1-y^2}\over\sqrt{1-x^2}}= c\)             [∵ log m - log n = log (m/n)]

\(\rm {\sqrt{1-y^2\over{1-x^2}}}= e^c = a\)

\(\rm {{1-y^2\over{1-x^2}}}= a^2 =C\)

y² - 1 = C(x² - 1)