250+ TOP MCQs on First Order Linear Differential Equations and Answers

Ordinary Differential Equations Questions and Answers for Freshers focuses on “First Order Linear Differential Equations”.

1. Solution of the differential equation (frac{dy}{dx}) + y cot ⁡x = cos⁡x is ______
a) (y cos ,x = frac{sin^2 x}{2} + c)
b) (y sin ,x = frac{sin^2 x}{2} + c)
c) (y sin ,x = frac{cos^2 x}{4} + c)
d) (y cos ,x = -frac{sin^2 x}{4} + c)
Answer: b
Explanation: (frac{dy}{dx}) + y cot⁡x = cos⁡x is of the form (frac{dy}{dx}) + Py = Q where P & Q is a function of x only
given DE is linear DE in y here P=cot x, Q=cos x, (e^{int P ,dx} = e^{int cot⁡x ,dx} = e^{log⁡sin⁡x} = sin x)
Linear DE solution is given by (y e^{int P ,dx} = int Q ,e^{int P ,dx} ,dx + c)
y sin x=∫(cos x*sin x) dx + c….. substitute sin x=t to solve integral
(y sin x = frac{t^2}{2} + c = frac{sin^2 x}{2} + c) is the solution.

2. For the differential equation (frac{dy}{dx}) – 3y cot⁡x = sin⁡2x; y=2 when x=(frac{pi}{2}), its particular solution is ______
a) y = 2cos² x + 4cos³ x
b) y = -2sin³ x + 4sin² x
c) y = -2sin² x + 4sin³ x
d) y = 4cos² x + 2sin³ x
Answer: c
Explanation: (frac{dy}{dx}) – 3y cot⁡x = sin⁡2x is of the form (frac{dy}{dx}) + Py = Q where P & Q is a function of x
only given DE is linear DE in y here P=-3cot x, Q=sin 2x, (e^{int P ,dx} = e^{int -3cot⁡x ,dx})
( =e^{-3 log⁡sin⁡x} = e^{log⁡(frac{1}{sin^3x})} = frac{1}{sin^3x})
Linear DE solution is given by (ye^{int P ,dx} = int Q ,e^{int P ,dx} dx + c)
(y frac{1}{sin^3x} = int sin ,2x frac{1}{sin^3x} ,dx + c = int 2 ,sin⁡x ,cos⁡x frac{1}{sin^3x} dx = int 2 ,cos⁡x frac{1}{sin^2x} dx)
substitute sin x=t to solve integral (y frac{1}{sin^3x} = int 2 frac{1}{t^2} ,dt + c = frac{-2}{t} + c = frac{-2}{sin⁡x} + c)
it is given that when x=π/2, y=2→2=-2+c→c=4 its particular solution is thus given
by (y frac{1}{sin^3x} = frac{-2}{sin⁡x} + 4 rightarrow y = -2sin^2 ,x + 4sin^3 ,x.)

3. Solution of the differential equation ((x+3y^2)frac{dy}{dx} = y(y>0)) is ______________
a) x=3y²+cy
b) y=2x²+c
c) x=2y²+(frac{c}{y})
d) y=3x³+c
Answer: a
Explanation: ((x+3y^2)frac{dy}{dx} = y) can be rearranged to (frac{dx}{dy} = frac{x}{y} + 3y rightarrow frac{dx}{dy} – frac{x}{y} = 3y )
above equation is of the form (frac{dx}{dy}) + Px = Q where P & Q is a function of y only
given DE is linear DE in x here( P = frac{-1}{y}, Q=3y, e^{int P ,dy} = e^{int frac{-1}{y} ,dy} = e^{log⁡y^{-1}} = frac{1}{y})
therefore its solution is given by (xe^{int P ,dy} = int Q ,e^{int P ,dy} dy + c)
(x frac{1}{y} = int 3y * frac{1}{y} ,dy + c = 3y + c)
i.e x=3y²+cy.

4. Solution of the differential equation (Big[frac{e^{-2sqrt{x}}}{sqrt{x}} – frac{y}{sqrt{x}}Big] frac{dy}{dx} = 1, ,x≠0) is _______
a) (ye^{2sqrt{x}}=2sqrt{x}+c)
b) (ye^{-2sqrt{x}}=2x^{-frac{3}{2}}+c)
c) (ye^{-2sqrt{x}}=3sqrt{x}+c)
d) (ye^{2sqrt{x}}=3x^{frac{3}{2}}+c)
Answer: a
Explanation: ( frac{1}{sqrt{x}} e^{-2sqrt{x}} – frac{1}{sqrt{x}} y = frac{dy}{dx} rightarrow frac{dy}{dx} + frac{1}{sqrt{x}} y = frac{1}{sqrt{x}} e^{-2sqrt{x}}) is of the form (frac{dy}{dx}) + Py = Q where P & Q is a function of x only.
given DE is linear DE in y here (P=frac{1}{sqrt{x}}, Q=frac{1}{sqrt{x}} e^{-2sqrt{x}})
(e^{int P ,dx} = e^{∫ frac{1}{sqrt{x}} ,dx} = e^{2sqrt{x}},)
Linear DE solution is given by (ye^{int P ,dx} = int Q ,e^{int P ,dx} dx + c)
(ye^{2sqrt{x}} = int frac{1}{sqrt{x}} e^{-2sqrt{x}} e^{2sqrt{x}} ,dx + c = 2sqrt{x} + c) is the solution.

5. Particular solution of the differential equation (frac{dy}{dx} = frac{x+y}{x}:y(1)=1) is _____
a) x =y log|x| + y
b) y = y log|x| + 2x
c) x = x log|y| + y
d) y = x log|x| + x
Answer: d
Explanation: (frac{dy}{dx} – frac{y}{x} = 1) is of the form (frac{dy}{dx}) + Py = Q where P & Q is a function of x only
given DE is linear DE in y here (P=frac{-1}{x}, Q=1, e^{int P ,dx} = e^{int frac{-1}{x} ,dx} = e^{logx^{-1}} = frac{1}{x})
Linear DE solution is given by (ye^{int P ,dx} = int Q ,e^{int P ,dx} dx + c rightarrow y frac{1}{x} = int 1 frac{1}{x} ,dx + c)
(frac{y}{x}) = log⁡|x|+c, given y(1)=1–>y=1 when x=1 i.e c=1 therefore its particular solution is given by y = x log|x| + x.

Global Education & Learning Series – Ordinary Differential Equations.

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