Chapter 9: Differential Equations — Online MCQ Test
MATHS · Grade 12 · CBSE(NCERT)
Practice Chapter 9: Differential Equations with a free chapter-wise online MCQ test.
This chapter covers: Core Definition: A differential equation is an equation that involves an independent variable, a dependent variable, and the derivatives of the dependent variable.Order of Equation....
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Chapter 9: Differential Equations — Important Questions & Answers
What is a differential equation?
- A. An equation involving an independent variable, dependent variable, and derivatives of the dependent variable
- B. An equation that only contains polynomial terms
- C. An equation with only algebraic expressions
- D. An equation without any variables
Answer: A. An equation involving an independent variable, dependent variable, and derivatives of the dependent variable
A differential equation is fundamentally defined as an equation involving independent variables, dependent variables, and the derivatives of the dependent variable.
A differential equation is fundamentally defined as an equation involving independent variables, dependent variables, and the derivatives of the dependent variable.
The order of the differential equation d³y/dx³ + 2(d²y/dx²)² + dy/dx = 0 is:
- A. 1
- B. 2
- C. 3
- D. 4
Answer: C. 3
The order of a differential equation is determined by the highest derivative present. Here, the highest derivative is d³y/dx³, which is third order.
The order of a differential equation is determined by the highest derivative present. Here, the highest derivative is d³y/dx³, which is third order.
Solve: dy/dx = e^(x-y)
- A. e^(-y) = e^x + C
- B. e^y = e^x + C
- C. y = e^x + C
- D. e^y + e^x = C
Answer: B. e^y = e^x + C
Rearranging: e^y dy = e^x dx. Integrating both sides: ∫e^y dy = ∫e^x dx gives e^y = e^x + C.
Rearranging: e^y dy = e^x dx. Integrating both sides: ∫e^y dy = ∫e^x dx gives e^y = e^x + C.
Solve dy/dx + y = e^(-x) with y(0) = 0:
- A. y = xe^(-x)
- B. y = x + e^(-x)
- C. y = e^x - 1
- D. y = (x + 1)e^(-x)
Answer: A. y = xe^(-x)
Here IF = e^(∫1 dx) = e^x. Solution: ye^x = ∫e^(-x)·e^x dx = ∫1 dx = x + C. With y(0) = 0: C = 0, so y = xe^(-x).
Here IF = e^(∫1 dx) = e^x. Solution: ye^x = ∫e^(-x)·e^x dx = ∫1 dx = x + C. With y(0) = 0: C = 0, so y = xe^(-x).
If a differential equation is both homogeneous and linear, which statement is true?
- A. It can only be solved by homogeneous method
- B. It can only be solved by linear method
- C. It can be solved by both methods, but they may give equivalent results
- D. It cannot be solved by either method
Answer: C. It can be solved by both methods, but they may give equivalent results
Some equations satisfy both conditions. While different approaches may appear distinct, they lead to equivalent general solutions when applied correctly.
Some equations satisfy both conditions. While different approaches may appear distinct, they lead to equivalent general solutions when applied correctly.