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Test: Differential Equations - 4 - JEE MCQ


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30 Questions MCQ Test Mathematics (Maths) for JEE Main & Advanced - Test: Differential Equations - 4

Test: Differential Equations - 4 for JEE 2025 is part of Mathematics (Maths) for JEE Main & Advanced preparation. The Test: Differential Equations - 4 questions and answers have been prepared according to the JEE exam syllabus.The Test: Differential Equations - 4 MCQs are made for JEE 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Differential Equations - 4 below.
Solutions of Test: Differential Equations - 4 questions in English are available as part of our Mathematics (Maths) for JEE Main & Advanced for JEE & Test: Differential Equations - 4 solutions in Hindi for Mathematics (Maths) for JEE Main & Advanced course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt Test: Differential Equations - 4 | 30 questions in 60 minutes | Mock test for JEE preparation | Free important questions MCQ to study Mathematics (Maths) for JEE Main & Advanced for JEE Exam | Download free PDF with solutions
Test: Differential Equations - 4 - Question 1
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Test: Differential Equations - 4 - Question 6
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A function y = f (x) satisfies the differential equation f (x) · sin 2x – cos x + (1 + sin2x) f ' (x) = 0 with initial condition y (0) = 0. The value of f (π/6) is equal to
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Test: Differential Equations - 4 - Question 9
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If for the differential equation ydx + y2dy = xdy, x ∈ R, y > 0 and y (1) = 1, then y (–3) is equal to
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Test: Differential Equations - 4 - Question 10
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Consider the differential equation

find the degree and the order of differential equation -
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Test: Differential Equations - 4 - Question 16
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The general solution of the differential equation (1 + tan y) (dx – dy) + 2xdy = 0 is -
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The general solution of the differential equation y (x2y + ex)dx – ex dy = 0 is -
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The curves satisfying the differential equation (1 – x2) y' + xy = ax are -
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Test: Differential Equations - 4 - Question 24
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If ϕ(x) is a differentiable function then the solution of dy + (yϕ'(x) – ϕ(x) ϕ'(x)) dx = 0 is
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Detailed Solution for Test: Differential Equations - 4 - Question 25
  1. Let p = dy/dx:
    This implies that d²y/dx² = dp/dx.
  2. Substitute into the original equation:
    y · dp/dx = p²
  3. Rewrite the equation using the chain rule:
    Since dp/dx = dp/dy · dy/dx = dp/dy · p, the equation becomes:
    y · p · dp/dy = p²
  4. Simplify the equation:
    Divide both sides by p (assuming p ≠ 0):
    dp/dy = p / y
  5. Separate the variables and integrate:
    dp/p = dy/y
    Integrate both sides:
    ln|p| = ln|y| + C₁
    Exponentiating both sides gives:
    p = C · y
    where C = e^C₁ is an arbitrary constant.
  6. Recall that p = dy/dx:
    So, dy/dx = C · y
  7. Solve the first-order differential equation:
    dy/y = C dx
    Integrate both sides:
    ln|y| = Cx + C₂
    Exponentiating gives:
    y = D · eCx
    where D = eC₂ is another arbitrary constant.
  8. Combine the constants:
    Since C and D are arbitrary constants, we can represent them as a single constant.
    Therefore, the general solution is:
    y = A · ekx
    where A and k are constants.
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Test: Differential Equations - 4 - Question 30
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xy = log y + C. Find y' = ? (where xy ≠ 1)
Detailed Solution for Test: Differential Equations - 4 - Question 30

xy = log y + C
=> d/dx (xy) = d/dx (log y)
=> y + x (dy/dx) = (1 / y) (dy/dx)
=> y + xy' = y' / y
=> y² + xy (y') = y'
=> (xy - 1) y' = -y²
=> y' = -y² / (1 - xy)

Thus, the given function is the solution of the differential equation.

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