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Which of the following are first order linear differential equations?

(A) dxdy+P1(y)x=Q1(y)\frac{dx}{dy} + P_1(y)x = Q_1(y) : P1(y)P_1(y) and Q1(y)Q_1(y) are functions of y or constant functions

(B) dydx+P2(x)y=Q2(x)\frac{dy}{dx} + P_2(x)y = Q_2(x) : P2(x)P_2(x) and Q2(x)Q_2(x) are functions of x or constant functions

(C) (x+y)dydx=x−2y(x + y)\frac{dy}{dx} = x - 2y

(D) (1+x2)dydx−2xy=x2+3(1 + x^2)\frac{dy}{dx} - 2xy = x^2 + 3

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 3

A first-order linear differential equation has the following characteristics:

  • Only first derivatives appear
  • The dependent variable and its derivative appear to the power of 1 only
  • Can be written in the form dydx+P(x)⋅y=Q(x)\frac{dy}{dx} + P(x) \cdot y = Q(x) or dxdy+P(y)⋅x=Q(y)\frac{dx}{dy} + P(y) \cdot x = Q(y)
  • PP and QQ must be functions of the independent variable only

Option (A): dxdy+P1(y)x=Q1(y)\frac{dx}{dy} + P_1(y)x = Q_1(y)

This is already in standard form where xx is the dependent variable and yy is independent.

The equation is first order and linear in xx.

Option (A) is a first-order linear differential equation.


Option (B): dydx+P2(x)y=Q2(x)\frac{dy}{dx} + P_2(x)y = Q_2(x)

This is already in standard form where yy is the dependent variable and xx is independent.

The equation is first order and linear in yy.

Option (B) is a first-order linear differential equation.


Option (C): (x+y)dydx=x−2y(x + y)\frac{dy}{dx} = x - 2y

Dividing both sides by (x+y)(x + y):

dydx=x−2yx+y\frac{dy}{dx} = \frac{x - 2y}{x + y}

dydx=xx+y−2yx+y\frac{dy}{dx} = \frac{x}{x+y} - \frac{2y}{x+y}

dydx+2yx+y=xx+y\frac{dy}{dx} + \frac{2y}{x+y} = \frac{x}{x+y}

The coefficient of yy is 2x+y\frac{2}{x+y}, which contains both xx and yy.

For a linear equation, the coefficient must be a function of xx only.

Option (C) is not a first-order linear differential equation.


Option (D): (1+x2)dydx−2xy=x2+3(1 + x^2)\frac{dy}{dx} - 2xy = x^2 + 3

(1+x2)dydx=2xy+x2+3(1 + x^2)\frac{dy}{dx} = 2xy + x^2 + 3

Dividing both sides by (1+x2)(1 + x^2):

dydx=2xy+x2+31+x2\frac{dy}{dx} = \frac{2xy + x^2 + 3}{1 + x^2}

dydx=2xy1+x2+x2+31+x2\frac{dy}{dx} = \frac{2xy}{1 + x^2} + \frac{x^2 + 3}{1 + x^2}

dydx−2x1+x2y=x2+31+x2\frac{dy}{dx} - \frac{2x}{1 + x^2}y = \frac{x^2 + 3}{1 + x^2}

Here P(x)=−2x1+x2P(x) = -\frac{2x}{1 + x^2} and Q(x)=x2+31+x2Q(x) = \frac{x^2 + 3}{1 + x^2} are both functions of xx only.

Option (D) is a first-order linear differential equation.


Options (A), (B), and (D) are first-order linear differential equations.

Option (C) is not a first-order linear differential equation.

The correct answer is: (A), (B) and (D) only

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