To find the particular solution to the differential equation [tex]\( x \frac{d y}{d x} + 3 y = \frac{\sin x}{x^2} \)[/tex] with the initial condition [tex]\( y\left(\frac{\pi}{2}\right) = 1 \)[/tex], let's follow these steps:
### Step 1: Transform the Differential Equation
First, we rewrite the equation in a standard form:
[tex]\[ y' + \frac{3}{x} y = \frac{\sin x}{x^3} \][/tex]
### Step 2: Identify the Integrating Factor
For a first-order linear differential equation of the form [tex]\( y' + P(x) y = Q(x) \)[/tex], the integrating factor [tex]\( \mu(x) \)[/tex] is given by:
[tex]\[ \mu(x) = e^{\int P(x) \, dx} \][/tex]
In this case, [tex]\( P(x) = \frac{3}{x} \)[/tex]:
[tex]\[ \mu(x) = e^{\int \frac{3}{x} \, dx} = e^{3 \ln |x|} = |x|^3 \][/tex]
Since we are considering [tex]\( x \neq 0 \)[/tex], we can write:
[tex]\[ \mu(x) = x^3 \][/tex]
### Step 3: Multiply Through by the Integrating Factor
Multiply the differential equation by the integrating factor [tex]\( x^3 \)[/tex]:
[tex]\[ x^3 y' + 3 x^2 y = \sin x \][/tex]
### Step 4: Simplify the Left Side
Notice that the left side of the equation is the derivative of [tex]\( x^3 y \)[/tex]:
[tex]\[ \frac{d}{dx} (x^3 y) = \sin x \][/tex]
### Step 5: Integrate Both Sides
Integrate both sides with respect to [tex]\( x \)[/tex]:
[tex]\[ x^3 y = \int \sin x \, dx \][/tex]
The integral of [tex]\( \sin x \)[/tex] is:
[tex]\[ \int \sin x \, dx = -\cos x + C \][/tex]
So, we have:
[tex]\[ x^3 y = -\cos x + C \][/tex]
[tex]\[ y = \frac{-\cos x + C}{x^3} \][/tex]
### Step 6: Apply the Initial Condition
Use the initial condition [tex]\( y\left(\frac{\pi}{2}\right) = 1 \)[/tex]:
[tex]\[ 1 = \frac{-\cos \left(\frac{\pi}{2}\right) + C}{\left(\frac{\pi}{2}\right)^3} \][/tex]
[tex]\[ 1 = \frac{0 + C}{\left(\frac{\pi}{2}\right)^3} \][/tex]
[tex]\[ 1 = \frac{C}{\left(\frac{\pi}{2}\right)^3} \][/tex]
Solve for [tex]\( C \)[/tex]:
[tex]\[ C = \left(\frac{\pi}{2}\right)^3 = \frac{\pi^3}{8} \][/tex]
### Step 7: Write the Particular Solution
Substitute [tex]\( C = \frac{\pi^3}{8} \)[/tex] back into the general solution:
[tex]\[ y = \frac{-\cos x + \frac{\pi^3}{8}}{x^3} \][/tex]
So the particular solution is:
[tex]\[ y = \frac{\frac{\pi^3}{8} - \cos x}{x^3} \][/tex]
None of the given options seems to match this solution directly. However, it can be rewritten to match Option B if we rewrite:
[tex]\[ y = \frac{\pi^3}{8 x^3} - \frac{\cos x}{x^3} \][/tex]
Which can be written as:
[tex]\[ y x^3 = \frac{\pi^3}{8} - \cos x \][/tex]
Thus the correct option is:
[tex]\[ \boxed{y x^3 + \cos x = \frac{\pi^3}{8}} \][/tex]
