250+ TOP MCQs on Bernoulli Equations and Answers

Ordinary Differential Equations Multiple Choice Questions on “Bernoulli Equations”.

1. Solution of the differential equation (frac{dy}{dx} + frac{y}{x} = y^2) x is ________
a) (frac{1}{y} = -x + c)
b) (frac{1}{xy} = -x + c)
c) (frac{1}{(xy)^2} = -y + c)
d) (frac{1}{xy} = -x^2 + c)
Answer: b
Explanation: Given equation is of the form (frac{dy}{dx} + Py = Qy^n) …divide by y²
where P and Q are functions of x hence this is Bernoulli’s equation in (y, frac{1}{y^2} frac{dy}{dx} + frac{1}{yx} = x)
put (frac{1}{y} = t rightarrow frac{-1}{y^2} frac{dy}{dx} = frac{dt}{dx}) substituting we get – (frac{dt}{dx} + frac{t}{x} = x ,or, frac{dt}{dx} – frac{t}{x} = -x)
this equation is linear in t i.e it is of the form
(frac{dt}{dx} + Pt = Q, I.F = e^{int P ,dx} =e^{int frac{-1}{x} ,dx} = e^{-log⁡x} = frac{1}{x})
its solution is
(te^{int P ,dx} = int Q e^{int P ,dx} ,dx + c rightarrow t frac{1}{x} = int -x * frac{1}{x} ,dx + c rightarrow frac{t}{x} = -x + c ,but, t = frac{1}{y})
thus solution is given by (frac{1}{xy} = -x + c).

2. Solution of the differential equation (frac{dy}{dx} -y ,tan⁡x = frac{sin⁡x cos^2⁡x}{y^2}) is ______
a) (y ,cos^2 x = frac{-cos^4 xsin^2 x}{2} + c)
b) (y^2 cos^2 x = frac{sin^6 x}{2} + c )
c) (y^3 cos^3 x = frac{-cos^6 x}{2} + c )
d) (y^4 cos^5 x = frac{sinx cos⁡x}{2} + c )
Answer: c
Explanation: (frac{dy}{dx} -y ,tan⁡x = frac{sin⁡x cos^2⁡x}{y^2}) multiplying by
(y^2 rightarrow y^2 frac{dy}{dx} – y^3 tan⁡x = sin⁡x ,cos^2⁡x )
put (t = y^3rightarrow3y^2 frac{dy}{dx} = frac{dt}{dx} ,or, y^2 frac{dy}{dx} = frac{1}{3} frac{dt}{dx})
substituting (frac{dt}{dx}-3t ,tan⁡x = 3sin⁡x ,cos^2⁡x )
this equation is linear in t i.e it is of the form (frac{dt}{dx} + Pt = Q, e^{int P ,dx} = e^{int -3tan⁡x ,dx})
(e^{-3 log⁡, sec⁡x} = cos^3 ,x) its solution is (te^{int P ,dx} = int Q ,e^{int P ,dx} dx + c)
t cos³ x=∫3sin⁡x cos²x cos³ x dx + c = ∫3sin⁡x cos⁵⁡x dx+c, put v=cos x
dv=-sin x dx i.e (int 3sin⁡x ,cos^5, ⁡x ,dx = int -3v^5 ,dx = frac{-v^6}{2} = frac{-cos^6 x}{2}) hence its solution becomes (t cos^3 x = frac{-cos^6 x}{2} + c rightarrow y^3 cos^3 ,x = frac{-cos^6 x}{2} + c.)

3. Solution of the differential equation 6y² dx – x(x³ + 2y)dy = 0 is ________
a) (frac{y}{x^3} = frac{-log⁡y}{2} + c )
b) (frac{y^2}{x^3} = frac{-log⁡x}{4} + c )
c) (frac{x}{y^3} = frac{-log⁡x}{2} + c )
d) (frac{x}{y^2} = frac{-log⁡y}{4} + c )
Answer: a
Explanation: Equation is reduced to (frac{dy}{dx} = frac{x(x^3+2y)}{6y^2} ,i.e, frac{dx}{dy}-frac{x}{3y} = frac{x^4}{6y^2}) …divide by x⁴
we get (frac{1}{x^4} frac{dy}{dx} – frac{1}{3x^3 y} = frac{1}{6y^2} ,put, frac{1}{x^3} = t rightarrow frac{-3}{x^4} frac{dx}{dy} = frac{dt}{dy} ,or, frac{1}{x^4} frac{dy}{dx} = frac{-1}{3} frac{dt}{dy} )
substituting (frac{-1}{3} frac{dt}{dy} – frac{t}{3y} = frac{1}{6y^2} ,or, frac{dt}{dy} + frac{t}{y} = frac{-1}{2y^2}) this equation is linear in t i.e it is of the form
(frac{dt}{dy} + Pt = Q ,where, P=frac{1}{y}, Q=frac{-1}{2y^2} ,I.F, = e^{int P ,dy} = e^{int frac{1}{y} ,dy} = e^{log⁡y} = y) its solution is (te^{int P ,dy} = int Q ,e^{int P ,dy} + c rightarrow ty = int y * frac{-1}{2y^2} ,dy + c = frac{-log⁡y}{2} + c)
(frac{y}{x^3} = frac{-log⁡y}{2} + c ) is the required solution.

Global Education & Learning Series – Ordinary Differential Equations.

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