Question 71 of 96intermediate🔧 ApplyLong Answer4 marks

For each of the differential equations given in Exercises 1 to 12, find the general solution:

(1+x2) dy+2xy dx=cot⁡x dx(x≠0)(1 + x^2) \, dy + 2xy \, dx = \cot x \, dx \quad (x \neq 0)
Correct Answer

y(1 + x²) = ln|sin x| + C

Exercise: EXERCISE 9.5 | Q: 8 | (Page 29)
For More Understanding

Explanation

This is a linear differential equation problem. The equation can be rearranged to the standard linear form dydx+Py=Q\dfrac{dy}{dx} + Py = Q, where the integrating factor method is applied.

The context shows similar linear differential equation solutions in Example 16, demonstrating the standard approach of finding:

I.F.=e∫P dx\text{I.F.} = e^{\int P \, dx}

and then solving:

y×I.F.=∫Q×I.F. dx+Cy \times \text{I.F.} = \int Q \times \text{I.F.} \, dx + C

Solution Steps

  1. Step 1: Rewrite the given equation:

    (1+x2) dy+2xy dx=cot⁡x dx(1 + x^2) \, dy + 2xy \, dx = \cot x \, dx

  2. Step 2: Divide by dxdx and rearrange:

    (1+x2) dydx+2xy=cot⁡x(1 + x^2) \, \frac{dy}{dx} + 2xy = \cot x

  3. Step 3: Convert to standard linear form:

    dydx+2x1+x2 y=cot⁡x1+x2\frac{dy}{dx} + \frac{2x}{1 + x^2} \, y = \frac{\cot x}{1 + x^2}

  4. Step 4: Identify PP and QQ:

    P=2x1+x2,Q=cot⁡x1+x2P = \frac{2x}{1 + x^2}, \qquad Q = \frac{\cot x}{1 + x^2}

  5. Step 5: Find the Integrating Factor:

    I.F.=e∫2x1+x2 dx=eln⁡(1+x2)=1+x2\text{I.F.} = e^{\int \frac{2x}{1 + x^2} \, dx} = e^{\ln(1 + x^2)} = 1 + x^2

  6. Step 6: Apply the solution formula:

    y(1+x2)=∫cot⁡x dx+Cy(1 + x^2) = \int \cot x \, dx + C

  7. Step 7: Integrate:

    ∫cot⁡x dx=ln⁡∣sin⁡x∣\int \cot x \, dx = \ln|\sin x|

  8. Step 8:

    Final Answer: y(1+x2)=ln⁡∣sin⁡x∣+C\boxed{\textbf{Final Answer: } y(1 + x^2) = \ln|\sin x| + C}
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