Question

The differential equation for which \[{\sin ^{ - 1}}x + {\sin ^{ - 1}}y = c\] is given by
A. \[\sqrt {1 - {x^2}} dx + \sqrt {1 - {y^2}} dy = 0\]
В. \[\sqrt {1 - {x^2}} dy + \sqrt {1 - {y^2}} dx = 0\]
C. \[\sqrt {1 - {x^2}} dy - \sqrt {1 - {y^2}} dx = 0\]
D. \[\sqrt {1 - {x^2}} dx - \sqrt {1 - {y^2}} dy = 0\]

Answer 1

Hint:
The general solution of a differential equation of the first order includes one arbitrary constant, whereas the general solution of a differential equation of the second order involves two arbitrary constants. The number of arbitrary constants present in the equation corresponding to the family of curves determines the order of a differential equation representing a family of curves.
Formula use:
\[y = {x^n}\]
Differentiate with respect to x
\[\frac{{dy}}{{dx}} = n{x^{n-1}}\]
\[{\sin ^{ - 1}}x= y\]
\[\frac{{dx}}{{\sqrt {1 - {x^2}} }}= dy\]
Complete step-by-step solution:
To create a differential equation from a given function, we differentiate it as many times as the number of arbitrary constants in the function and then eliminate the arbitrary constants.
We have been given the equation in the question as,
 \[{\sin ^{ - 1}}x + {\sin ^{ - 1}}y = c\]------ (1)
Differentiation is linear.
We can differentiate summands separately and pull out constants factor.
By differentiating the equation (1) on both sides with respect to \[{\rm{x}}\], we get
\[\frac{{dx}}{{\sqrt {1 - {x^2}} }} + \frac{{dy}}{{\sqrt {1 - {y^2}} }} = 0\]
The above equation can be rewritten using differentiation identities as,
\[ \Rightarrow \sqrt {1 - {x^2}} dy + \sqrt {1 - {y^2}} dx = 0\]
Therefore, the differential equation for which \[{\sin ^{ - 1}}x + {\sin ^{ - 1}}y = c\] is given by\[\sqrt {1 - {x^2}} dy + \sqrt {1 - {y^2}} dx = 0\]
Hence, the option B is correct.
Note:
When solving differential equation problems, students sometimes fail to remove the arbitrary constants that are variable coefficients, resulting in a completely different output. As a result, ensure that all arbitrary constants or parameters are removed from differential equations. Students frequently write order instead of degree and vice versa. Before proceeding further in the chapter, it is critical to fully comprehend the two. Remember that the specific solution is one in which the integral constant has a numerical value.