Unit 8: Differential Equations — Online MCQ Test
MATHEMATICS - 2B · CLASS 12 INTERMEDIATE 2 YEAR · Telangana State Board
Practice Unit 8: Differential Equations with a free chapter-wise online MCQ test.
This chapter covers: The final chapter details order degree of differential equations formation of differential equations solutions of variable separable homogeneous and linear first-order differential....
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Unit 8: Differential Equations — Important Questions & Answers
What is the order of the differential equation \(\frac{d^2y}{dx^2} + 3\frac{dy}{dx} - 5y = 0\)?
- A. 1
- B. 2
- C. 3
- D. 5
Answer: B. 2
The highest derivative present is \(\frac{d^2y}{dx^2}\), so the order is 2.
The highest derivative present is \(\frac{d^2y}{dx^2}\), so the order is 2.
Which of the following is a differential equation?
- A. \(y = x^2 + 1\)
- B. \(\frac{dy}{dx} = x + y\)
- C. \(x^2 + y^2 = 1\)
- D. \(y = \sin x\)
Answer: B. \(\frac{dy}{dx} = x + y\)
A differential equation contains derivatives of the dependent variable. Only option B includes a derivative.
A differential equation contains derivatives of the dependent variable. Only option B includes a derivative.
Solve the separable differential equation \(\frac{dy}{dx} = xy\).
- A. \(y = Ce^{x^2/2}\)
- B. \(y = Ce^{2x}\)
- C. \(y = Cx^2\)
- D. \(y = C\ln x\)
Answer: A. \(y = Ce^{x^2/2}\)
Separating variables gives \(\frac{dy}{y} = x\,dx\). Integrating, \(\ln|y| = \frac{x^2}{2} + C\), hence \(y = Ce^{x^2/2}\).
Separating variables gives \(\frac{dy}{y} = x\,dx\). Integrating, \(\ln|y| = \frac{x^2}{2} + C\), hence \(y = Ce^{x^2/2}\).
If the differential equation is \(\frac{dy}{dx} = \frac{x+y}{x}\), then after substituting \(y = vx\), the equation becomes
- A. \(v + x\frac{dv}{dx} = 1 + v\)
- B. \(x\frac{dv}{dx} = 1\)
- C. \(x\frac{dv}{dx} = v\)
- D. \(\frac{dv}{dx} = x+v\)
Answer: B. \(x\frac{dv}{dx} = 1\)
Using \(y=vx\), we get \(\frac{dy}{dx}=v+x\frac{dv}{dx}\). Substituting gives \(v+x\frac{dv}{dx}=1+v\), so \(x\frac{dv}{dx}=1\).
Using \(y=vx\), we get \(\frac{dy}{dx}=v+x\frac{dv}{dx}\). Substituting gives \(v+x\frac{dv}{dx}=1+v\), so \(x\frac{dv}{dx}=1\).
If a first-order differential equation has integrating factor \(e^{\int 2x\,dx}\), then the corresponding coefficient \(P(x)\) is
- A. x
- B. 2x
- C. x^2
- D. 2
Answer: B. 2x
The integrating factor for \(\frac{dy}{dx} + P(x)y = Q(x)\) is \(e^{\int P(x)dx}\). Hence \(P(x)=2x\).
The integrating factor for \(\frac{dy}{dx} + P(x)y = Q(x)\) is \(e^{\int P(x)dx}\). Hence \(P(x)=2x\).