JEE Exam > JEE Questions > The equation of the curve satisfying the diff...
Start Learning for Free
The equation of the curve satisfying the differential equation y(x y3 ) dx = x (y3 – x) dy and passing through the point (1, 1) is (A) y3 – 2x 3x2 y = 0 (B) y3 2x 3x2 y = 0 (C) y3 2x – 3x2 y = 0 (D) none of these?
Most Upvoted Answer
The equation of the curve satisfying the differential equation y(x y...
Given:
The differential equation is: y(x + y^3)dx = x(y^3 - x)dy
The curve passes through the point (1, 1)
To find:
The equation of the curve.
Solution:
Step 1: Separating Variables
Let's start by separating the variables in the given differential equation.
y(x + y^3)dx = x(y^3 - x)dy
Divide both sides by xy(x - y^3):
(1 + y^3/x)dx = (y^3 - x)/ydy
Step 2: Integrating
Next, we integrate both sides of the equation.
∫(1 + y^3/x)dx = ∫(y^3 - x)/ydy
Integrating the left side:
∫(1 + y^3/x)dx = x + y^3∫(1/x)dx = x + y^3ln|x| + C1, where C1 is the constant of integration.
Integrating the right side:
∫(y^3 - x)/ydy = ∫(y^2 - x/y)dy = y^3/3 - xy + C2, where C2 is the constant of integration.
Step 3: Combining the Integrals
Now, we can combine the integrals and simplify the equation.
x + y^3ln|x| + C1 = y^3/3 - xy + C2
Rearranging the terms:
x + xy - y^3/3 - y^3ln|x| = C2 - C1
Combining the constants:
C = C2 - C1, where C is the constant of integration.
Simplifying further:
x(1 + y) - (y^3/3)(1 + ln|x|) = C
Step 4: Applying the Initial Condition
The curve passes through the point (1, 1), so we can substitute these values into the equation to find the specific value of the constant C.
1(1 + 1) - (1^3/3)(1 + ln|1|) = C
2 - 1/3 = C
C = 5/3
Step 5: Final Equation
Substituting the value of C back into the equation, we get:
x(1 + y) - (y^3/3)(1 + ln|x|) = 5/3
Simplifying further, we arrive at the equation of the curve:
3x(1 + y) - y^3(1 + ln|x|) = 5
Therefore, the correct answer is (D) none of these. The equation of the curve satisfying the given differential equation and passing through the point (1, 1) is 3x(1 + y) - y^3(1 + ln|x|) = 5.
The differential equation is: y(x + y^3)dx = x(y^3 - x)dy
The curve passes through the point (1, 1)
To find:
The equation of the curve.
Solution:
Step 1: Separating Variables
Let's start by separating the variables in the given differential equation.
y(x + y^3)dx = x(y^3 - x)dy
Divide both sides by xy(x - y^3):
(1 + y^3/x)dx = (y^3 - x)/ydy
Step 2: Integrating
Next, we integrate both sides of the equation.
∫(1 + y^3/x)dx = ∫(y^3 - x)/ydy
Integrating the left side:
∫(1 + y^3/x)dx = x + y^3∫(1/x)dx = x + y^3ln|x| + C1, where C1 is the constant of integration.
Integrating the right side:
∫(y^3 - x)/ydy = ∫(y^2 - x/y)dy = y^3/3 - xy + C2, where C2 is the constant of integration.
Step 3: Combining the Integrals
Now, we can combine the integrals and simplify the equation.
x + y^3ln|x| + C1 = y^3/3 - xy + C2
Rearranging the terms:
x + xy - y^3/3 - y^3ln|x| = C2 - C1
Combining the constants:
C = C2 - C1, where C is the constant of integration.
Simplifying further:
x(1 + y) - (y^3/3)(1 + ln|x|) = C
Step 4: Applying the Initial Condition
The curve passes through the point (1, 1), so we can substitute these values into the equation to find the specific value of the constant C.
1(1 + 1) - (1^3/3)(1 + ln|1|) = C
2 - 1/3 = C
C = 5/3
Step 5: Final Equation
Substituting the value of C back into the equation, we get:
x(1 + y) - (y^3/3)(1 + ln|x|) = 5/3
Simplifying further, we arrive at the equation of the curve:
3x(1 + y) - y^3(1 + ln|x|) = 5
Therefore, the correct answer is (D) none of these. The equation of the curve satisfying the given differential equation and passing through the point (1, 1) is 3x(1 + y) - y^3(1 + ln|x|) = 5.
Attention JEE Students!
To make sure you are not studying endlessly, EduRev has designed JEE study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in JEE.
|
Explore Courses for JEE exam
|
|
Similar JEE Doubts
Top Courses for JEEView all
The equation of the curve satisfying the differential equation y(x y3 ) dx = x (y3 – x) dy and passing through the point (1, 1) is (A) y3 – 2x 3x2 y = 0 (B) y3 2x 3x2 y = 0 (C) y3 2x – 3x2 y = 0 (D) none of these?
Question Description
The equation of the curve satisfying the differential equation y(x y3 ) dx = x (y3 – x) dy and passing through the point (1, 1) is (A) y3 – 2x 3x2 y = 0 (B) y3 2x 3x2 y = 0 (C) y3 2x – 3x2 y = 0 (D) none of these? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The equation of the curve satisfying the differential equation y(x y3 ) dx = x (y3 – x) dy and passing through the point (1, 1) is (A) y3 – 2x 3x2 y = 0 (B) y3 2x 3x2 y = 0 (C) y3 2x – 3x2 y = 0 (D) none of these? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The equation of the curve satisfying the differential equation y(x y3 ) dx = x (y3 – x) dy and passing through the point (1, 1) is (A) y3 – 2x 3x2 y = 0 (B) y3 2x 3x2 y = 0 (C) y3 2x – 3x2 y = 0 (D) none of these?.
The equation of the curve satisfying the differential equation y(x y3 ) dx = x (y3 – x) dy and passing through the point (1, 1) is (A) y3 – 2x 3x2 y = 0 (B) y3 2x 3x2 y = 0 (C) y3 2x – 3x2 y = 0 (D) none of these? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The equation of the curve satisfying the differential equation y(x y3 ) dx = x (y3 – x) dy and passing through the point (1, 1) is (A) y3 – 2x 3x2 y = 0 (B) y3 2x 3x2 y = 0 (C) y3 2x – 3x2 y = 0 (D) none of these? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The equation of the curve satisfying the differential equation y(x y3 ) dx = x (y3 – x) dy and passing through the point (1, 1) is (A) y3 – 2x 3x2 y = 0 (B) y3 2x 3x2 y = 0 (C) y3 2x – 3x2 y = 0 (D) none of these?.
Solutions for The equation of the curve satisfying the differential equation y(x y3 ) dx = x (y3 – x) dy and passing through the point (1, 1) is (A) y3 – 2x 3x2 y = 0 (B) y3 2x 3x2 y = 0 (C) y3 2x – 3x2 y = 0 (D) none of these? in English & in Hindi are available as part of our courses for JEE.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of The equation of the curve satisfying the differential equation y(x y3 ) dx = x (y3 – x) dy and passing through the point (1, 1) is (A) y3 – 2x 3x2 y = 0 (B) y3 2x 3x2 y = 0 (C) y3 2x – 3x2 y = 0 (D) none of these? defined & explained in the simplest way possible. Besides giving the explanation of
The equation of the curve satisfying the differential equation y(x y3 ) dx = x (y3 – x) dy and passing through the point (1, 1) is (A) y3 – 2x 3x2 y = 0 (B) y3 2x 3x2 y = 0 (C) y3 2x – 3x2 y = 0 (D) none of these?, a detailed solution for The equation of the curve satisfying the differential equation y(x y3 ) dx = x (y3 – x) dy and passing through the point (1, 1) is (A) y3 – 2x 3x2 y = 0 (B) y3 2x 3x2 y = 0 (C) y3 2x – 3x2 y = 0 (D) none of these? has been provided alongside types of The equation of the curve satisfying the differential equation y(x y3 ) dx = x (y3 – x) dy and passing through the point (1, 1) is (A) y3 – 2x 3x2 y = 0 (B) y3 2x 3x2 y = 0 (C) y3 2x – 3x2 y = 0 (D) none of these? theory, EduRev gives you an
ample number of questions to practice The equation of the curve satisfying the differential equation y(x y3 ) dx = x (y3 – x) dy and passing through the point (1, 1) is (A) y3 – 2x 3x2 y = 0 (B) y3 2x 3x2 y = 0 (C) y3 2x – 3x2 y = 0 (D) none of these? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup