Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Differential Equations

Question:

Match List – I with List – II.

LIST I

LIST II

 A. $\frac{d y}{d x}=e^{x-y}+x^2 e^{-y}$

 I. Linear Differential Equation of type $\frac{d y}{d x}+P y=Q$

 B. $\frac{d y}{d x}-2 y=\cos 3 x$

 II. Homogeneous Differential Equation

 C. $x^2 \frac{d y}{d x}=x^2-2 y^2+x y$

 III. Linear Differential Equation of type $\frac{d y}{d x}+P x=Q$

 D. $\left(x+2 y^3\right) \frac{d y}{d x}=y$

 IV. Variable separable

Choose the correct answer from the options given below:

Options:

A-II, B-III, C-IV, D-I

A-IV, B-III, C-II, D-I

A-IV, B-I, C-II, D-III

A-III, B-I, C-IV, D-II

Correct Answer:

A-IV, B-I, C-II, D-III

Explanation:

The correct answer is Option (3) - A-IV, B-I, C-II, D-III

(A) $\frac{d y}{d x}=e^{x-y}+x^2 e^{-y}⇒\frac{d y}{d x}=\frac{e^x+x^2}{e^y}$

so $\int e^ydy=\int e^x+x^2dx$ (IV)

(B) $\frac{d y}{d x}-2 y=\cos 3 x$ (I)

(C) $x^2 \frac{d y}{d x}=x^2-2 y^2+x y⇒\frac{d y}{d x}=1-2(\frac{y}{x})^2+(\frac{y}{x})=f(\frac{y}{x})$ (II)

(D) $\left(x+2 y^3\right) \frac{d y}{d x}=y$

so $\frac{dx}{dy}-\frac{x}{y}=2y^2$ (III)