Question 35 of 96intermediate🔧 ApplyLong Answer4 marks

For each of the differential equations in Exercises 11 to 14, find a particular solution satisfying the given condition:

(x3+x2+x+1)dydx=2x2+x;y=1 when x=0(x^3 + x^2 + x + 1) \frac{dy}{dx} = 2x^2 + x; \quad y = 1 \text{ when } x = 0
Correct Answer
y=34ln⁡(x2+1)+12ln⁡∣x+1∣−12tan⁡−1(x)+1y = \frac{3}{4} \ln(x^2+1) + \frac{1}{2} \ln|x+1| - \frac{1}{2} \tan^{-1}(x) + 1
Exercise: EXERCISE 9.3 | Q: 11 | (Page 12)
For More Understanding

Explanation

This is a separable differential equation. The key steps involve: (1) separating variables, (2) factoring the denominator using grouping, (3) applying partial fractions decomposition, (4) integrating each term, and (5) applying the initial condition y=1y = 1 when x=0x = 0 to find the constant of integration.

The denominator factors as (x2+1)(x+1)(x^2+1)(x+1), and the partial fractions yield terms that integrate to logarithmic and inverse trigonometric functions.

Solution Steps

  1. Step 1: Separate variables:

    dy=2x2+xx3+x2+x+1 dxdy = \frac{2x^2 + x}{x^3 + x^2 + x + 1} \, dx

  2. Step 2: Factor denominator:

    x3+x2+x+1=(x2+1)(x+1)x^3 + x^2 + x + 1 = (x^2 + 1)(x + 1)

  3. Step 3: Apply partial fractions:

    2x2+x(x2+1)(x+1)=3x/2−1/2x2+1+1/2x+1\frac{2x^2 + x}{(x^2+1)(x+1)} = \frac{3x/2 - 1/2}{x^2+1} + \frac{1/2}{x+1}

  4. Step 4: Integrate:

    y=34ln⁡(x2+1)−12tan⁡−1(x)+12ln⁡∣x+1∣+Cy = \frac{3}{4} \ln(x^2+1) - \frac{1}{2} \tan^{-1}(x) + \frac{1}{2} \ln|x+1| + C

  5. Step 5: Apply condition y=1y = 1 when x=0x = 0:

    1=0−0+0+C,so C=11 = 0 - 0 + 0 + C, \quad \text{so } C = 1

  6. Step 6: Final particular solution:

    y=34ln⁡(x2+1)+12ln⁡∣x+1∣−12tan⁡−1(x)+1y = \frac{3}{4} \ln(x^2+1) + \frac{1}{2} \ln|x+1| - \frac{1}{2} \tan^{-1}(x) + 1
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