Differential Equations and Modeling
CBSE · Class 12 · Applied Mathematics
Most important questions from Differential Equations and Modeling for CBSE Class 12 Applied Mathematics board exam 2026. MCQs, short answer, and long answer questions with marks.
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Sample Questions
Which of the following are solutions to dy/dx + y = 0? (Select all correct answers)
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y = e^(-x), y = 2e^(-x), y = -3e^(-x), y = ce^(-x)
Step 1: Solve dy/dx = -y by separation: dy/y = -dx. Step 2: Integrate: ln|y| = -x + C. Step 3: y = Ae^(-x) where A is any constant. All options with form Ae^(-x) are correct, while e^x gives dy/dx = e^x ≠ -y.
A population grows according to dP/dt = kP. If P(0) = 100 and P(2) = 200, find k.
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k = (ln 2)/2
Step 1: Solve dP/dt = kP: P = P₀e^(kt). Step 2: Using P(0) = 100: P = 100e^(kt). Step 3: Using P(2) = 200: 200 = 100e^(2k). Step 4: Solve: 2 = e^(2k), so 2k = ln 2, therefore k = (ln 2)/2.
The differential equation y'' + 4y = 0 has which order and degree?
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Order = 2, Degree = 1
Step 1: Identify highest derivative: y'' = d²y/dx² (second order). Step 2: Check polynomial form: equation is polynomial in derivatives. Step 3: Find power of highest derivative: (y'')¹, so degree = 1. Therefore: Order = 2, Degree = 1.
Solve: dy/dx = y/x, x > 0
Show answer
y = cx
Step 1: Separate variables: dy/y = dx/x. Step 2: Integrate both sides: ∫dy/y = ∫dx/x. Step 3: ln|y| = ln|x| + ln|c|. Step 4: ln|y| = ln|cx|, so y = cx.
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- NCERT Official — ncert.nic.in
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- CBSE Official — cbse.gov.in
- National Education Policy 2020 — education.gov.in
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