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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?
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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.
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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?
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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?.
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