Question

Find the solution to the equation \[{\sin ^{ - 1}}\left( {\dfrac{{dy}}{{dx}}} \right) = x + y\].
A. \[\tan \left( {x + y} \right) + \sec \left( {x + y} \right) = x + c\]
B. \[\tan \left( {x + y} \right) - \sec \left( {x + y} \right) = x + c\]
C. \[\tan \left( {x + y} \right) + \sec \left( {x + y} \right) + x + c = 0\]
D. None of these

Answer 1

Hint: To solve the differential equation, we will use the substitution method. We equate \[x + y\] with z and differentiate it with respect to x. Substitute \[x + y = z\] and its differentiation in the given equation and solve it by using integration formulas.

Formula used:
Trigonometry identity:
\[1 - {\sin ^2}\theta = {\cos ^2}\theta \]
Integration formula:
\[\int {{{\sec }^2}\theta d\theta } = \tan \theta + c\]
\[\int {\sec \theta \tan \theta d\theta } = \sec \theta + c\]

Complete step by step solution:
Given differential equation:
\[{\sin ^{ - 1}}\left( {\dfrac{{dy}}{{dx}}} \right) = x + y\]
Rewrite the equation:
\[\dfrac{{dy}}{{dx}} = \sin \left( {x + y} \right)\] …..(i)
Assume that \[x + y = z\]
Differentiate with respect to x
\[\dfrac{d}{{dx}}\left( {x + y} \right) = \dfrac{{dz}}{{dx}}\]
\[ \Rightarrow 1 + \dfrac{{dy}}{{dx}} = \dfrac{{dz}}{{dx}}\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{dz}}{{dx}} - 1\]
Substitute \[x + y = z\] and \[\dfrac{{dy}}{{dx}} = \dfrac{{dz}}{{dx}} - 1\] in equation (i)
\[\dfrac{{dz}}{{dx}} - 1 = \sin z\]
\[ \Rightarrow \dfrac{{dz}}{{dx}} = \sin z + 1\]
Divide both sides by \[1 + \sin z\]
\[ \Rightarrow \dfrac{{dz}}{{\sin z + 1}} = dx\]
Multiply \[1 - \sin z\] with denominator and numerator of left side expression
\[ \Rightarrow \dfrac{{\left( {1 - \sin z} \right)dz}}{{\left( {1 - \sin z} \right)\left( {\sin z + 1} \right)}} = dx\]
Apply the algebraic identity \[\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}\]
 \[ \Rightarrow \dfrac{{\left( {1 - \sin z} \right)dz}}{{1 - {{\sin }^2}z}} = dx\]
Now using the trigonometric identity \[1 - {\sin ^2}\theta = {\cos ^2}\theta \]
\[ \Rightarrow \dfrac{{\left( {1 - \sin z} \right)dz}}{{{{\cos }^2}z}} = dx\]
\[ \Rightarrow \dfrac{1}{{{{\cos }^2}z}}dz - \dfrac{{\sin z}}{{{{\cos }^2}z}}dz = dx\]
\[ \Rightarrow {\sec ^2}zdz - \tan z\sec zdz = dx\]
Taking integration both sides of the equation:
\[ \Rightarrow \int {{{\sec }^2}zdz} - \int {\tan z\sec zdz} = \int {dx} \]
Applying integration formula
\[ \Rightarrow \tan z - \sec z = x + c\]
Now putting the value of z
\[ \Rightarrow \tan \left( {x + y} \right) - \sec \left( {x + y} \right) = x + c\]
Hence option B is the correct option.

Note: There is another answer to the question.
\[{\sin ^{ - 1}}\left( {\dfrac{{dy}}{{dx}}} \right) = x + y\]
Rewrite the equation:
\[\dfrac{{dy}}{{dx}} = \sin \left( {x + y} \right)\] …..(i)
Assume that \[x + y = z\]
Differentiate with respect to x
\[\dfrac{d}{{dx}}\left( {x + y} \right) = \dfrac{{dz}}{{dx}}\]
\[ \Rightarrow 1 + \dfrac{{dy}}{{dx}} = \dfrac{{dz}}{{dx}}\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{dz}}{{dx}} - 1\]
Substitute \[x + y = z\] and \[\dfrac{{dy}}{{dx}} = \dfrac{{dz}}{{dx}} - 1\] in equation (i)
\[\dfrac{{dz}}{{dx}} - 1 = \sin z\]
\[ \Rightarrow \dfrac{{dz}}{{dx}} = \sin z + 1\]
Divide both sides by \[1 + \sin z\]
\[ \Rightarrow \dfrac{{dz}}{{\sin z + 1}} = dx\]
\[ \Rightarrow \dfrac{{dz}}{{{{\left( {\sin \dfrac{z}{2} + \cos \dfrac{z}{2}} \right)}^2}}} = dx\]
\[ \Rightarrow \dfrac{{{{\sec }^2}\dfrac{z}{2}dz}}{{{{\left( {\tan \dfrac{z}{2} + 1} \right)}^2}}} = dx\]
Assume that \[1 + \tan \dfrac{z}{2} = t\]\[ \Rightarrow \dfrac{1}{2}{\sec ^2}\dfrac{z}{2}dz = dt\]
\[ \Rightarrow \dfrac{{2dt}}{{{t^2}}} = dx\]
Integrating both sides
\[ \Rightarrow - \dfrac{2}{t} = x + c\]
Putting \[1 + \tan \dfrac{z}{2} = t\]
\[ \Rightarrow - \dfrac{2}{{1 + \tan \dfrac{z}{2}}} = x + c\]
Putting z = x + y
\[ \Rightarrow - \dfrac{2}{{1 + \tan \dfrac{{x + y}}{2}}} = x + c\]
\[ \Rightarrow \left( {x + c} \right)\left( {1 + \tan \dfrac{{x + y}}{2}} \right) + 2 = 0\]