For each of the differential equations given in Exercises 13 to 15,…

Question 77 of 96intermediate🔧 ApplyLong Answer4 marks

For each of the differential equations given in Exercises 13 to 15, find a particular solution satisfying the given condition:

(1 + x^2) \frac{dy}{dx} + 2xy = \frac{1}{1+x^2}; \quad y = 0 \text{ when } x = 1

Correct Answer

y(1 + x²) = tan⁻¹(x) - π/4

Exercise: EXERCISE 9.5 | Q: 14 | (Page 30)

For More Understanding

Explanation

This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$ . The student needs to identify this form, find the integrating factor, integrate, and then apply the initial condition to find the constant of integration.

Solution Steps

Step 1: Rewrite the equation in standard linear form. Dividing both sides by $(1 + x^2)$ :
$\frac{dy}{dx} + \frac{2x}{1+x^2}y = \frac{1}{(1+x^2)^2}$
Step 2: Identify $P = \frac{2x}{1+x^2}$ and $Q = \frac{1}{(1+x^2)^2}$ . This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$ .
Step 3: Find the integrating factor (IF).
$\text{IF} = e^{\int P \, dx} = e^{\int \frac{2x}{1+x^2} \, dx}$
Let $t = 1 + x^2$ , so $dt = 2x \, dx$ . Thus:
$\int \frac{2x}{1+x^2} \, dx = \int \frac{dt}{t} = \ln|t| = \ln(1+x^2)$
Therefore,
$\text{IF} = e^{\ln(1+x^2)} = 1 + x^2$
Step 4: Apply the integrating factor formula.
$y(\text{IF}) = \int Q(\text{IF}) \, dx + C$
So,
$y(1+x^2) = \int \frac{1}{(1+x^2)^2}(1+x^2) \, dx + C = \int \frac{1}{1+x^2} \, dx + C = \tan^{-1}(x) + C$
Step 5: Apply the initial condition $y = 0$ when $x = 1$ . Substituting:
$0(1+1) = \tan^{-1}(1) + C$
So,
$0 = \frac{\pi}{4} + C$
giving $C = -\frac{\pi}{4}$ .
Step 6: Write the particular solution:
$y(1+x^2) = \tan^{-1}(x) - \frac{\pi}{4}$
or equivalently,
$y = \frac{\tan^{-1}(x) - \frac{\pi}{4}}{1+x^2}$
∴ The particular solution is $y = \dfrac{\tan^{-1}(x) - \pi/4}{1+x^2}$ .

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