For each of the differential equations given in Exercises 1 to 12,…

Question 72 of 96intermediate🔧 ApplyLong Answer4 marks

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

x \frac{dy}{dx} + y - x + xy \cot x = 0 \quad (x \neq 0)

Correct Answer

The general solution is:

y \, x \sin x = -x \cos x + \sin x + C

or equivalently,

y = -\cot x + \frac{1}{x} + \frac{C}{x \sin x}

Exercise: EXERCISE 9.5 | Q: 9 | (Page 30)

For More Understanding

Explanation

This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$ . After rearranging the given equation and dividing by $x$ , we identify $P = \left(\frac{1}{x} + \cot x\right)$ and $Q = 1$ .

The integrating factor is calculated using $e^{\int P \, dx}$ , which gives $x \sin x$ . Using the standard formula for solving linear differential equations, we integrate $Q \times \text{I.F.}$ to get the final solution.

Solution Steps

Step 1: Rewrite the given equation:
$x \frac{dy}{dx} + y - x + xy \cot x = 0$
Dividing by $x$ (since $x \neq 0$ ):
$\frac{dy}{dx} + \frac{y}{x} + y \cot x = 1$
Step 2: Rearrange to standard linear form:
$\frac{dy}{dx} + y\left(\frac{1}{x} + \cot x\right) = 1$
Here $P = \frac{1}{x} + \cot x$ and $Q = 1$
Step 3: Calculate the Integrating Factor (I.F.):
$I.F. = e^{\int P \, dx} = e^{\int \left(\frac{1}{x} + \cot x\right) dx} = e^{\log|x| + \log|\sin x|} = e^{\log|x \sin x|} = x \sin x$
Step 4: Apply the general solution formula:
$y \times I.F. = \int Q \times I.F. \, dx + C$
Step 5: Calculate:
$y \cdot x \sin x = \int x \sin x \, dx + C$
Using integration by parts:
$\int x \sin x \, dx = -x \cos x + \sin x$
Step 6: Therefore, the general solution is:
$y \cdot x \sin x = -x \cos x + \sin x + C$
or
$y = -\cot x + \frac{1}{x} + \frac{C}{x \sin x}$

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