Question 96 of 96intermediate🔧 ApplyMCQ1 mark

The general solution of the differential equation exdy+(yex+2x)dx=0e^x dy + (y e^x + 2x) dx = 0 is

Exercise: Miscellaneous Exercise on Chapter 9 | Q: 15 | (Page 36)
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Explanation

The correct answer is Option (C). The given differential equation can be rewritten as exe^{x}(dy/dx) + y·exe^{x} + 2x = 0, which simplifies to dy/dx + y = -2xe^(-x). This is a linear differential equation of the form dy/dx + Py = Q, where P = 1. The integrating factor is e^(∫1 dx) = exe^{x}. Multiplying both sides by the integrating factor gives d/dx(y·exe^{x}) = -2x. Integrating both sides yields y·exe^{x} = -x² + C, which can be written as y·exe^{x} + x² = C. This matches option (C).

Solution Steps

  1. Step 1: Rewrite the equation as exe^{x}(dy/dx) + y·exe^{x} + 2x = 0

  2. Step 2: Simplify to dy/dx + y = -2xe^(-x)

  3. Step 3: Identify this as a linear differential equation with P = 1

  4. Step 4: Find the integrating factor: I.F. = e^(∫1 dx) = exe^{x}

  5. Step 5: Multiply by I.F.: d/dx(y·exe^{x}) = -2x

  6. Step 6: Integrate both sides: y·exe^{x} = -x² + C

  7. Step 7: Rearrange to get y·exe^{x} + x² = C

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