
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
162.6k+ views
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
\[ \Rightarrow - \dfrac{2}{{1 + \tan \dfrac{{x + y}}{2}}} = x + c\]
\[ \Rightarrow \left( {x + c} \right)\left( {1 + \tan \dfrac{{x + y}}{2}} \right) + 2 = 0\]
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
\[ \Rightarrow - \dfrac{2}{{1 + \tan \dfrac{{x + y}}{2}}} = x + c\]
\[ \Rightarrow \left( {x + c} \right)\left( {1 + \tan \dfrac{{x + y}}{2}} \right) + 2 = 0\]
Recently Updated Pages
JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NEET 2025 – Every New Update You Need to Know

Verb Forms Guide: V1, V2, V3, V4, V5 Explained

NEET Total Marks 2025

1 Billion in Rupees
