`xy'-y=x^2(xcosx+sinx)` and if `f(pi)=pi` then find `f''(pi/2)+f(pi/2)=`

Answer

To solve the given differential equation \( xy' - y = x^2(x \cos x + \sin x) \) with the condition \( f(\pi) = \pi \), we need to find \( f''(\frac{\pi}{2}) + f(\frac{\pi}{2}) \).

### Step 1: Rewrite the equation
We start by rewriting the equation:
\[ xy' - y = x^2(x \cos x + \sin x) \]
Dividing both sides by \( x \):
\[ y' - \frac{y}{x} = x(x \cos x + \sin x) \]

### Step 2: Identify \( p \) and \( q \)
This is a first-order linear ordinary differential equation of the form:
\[ y' + p y = q \]
where \( p = -\frac{1}{x} \) and \( q = x(x \cos x + \sin x) \).

### Step 3: Find the integrating factor
The integrating factor \( \mu(x) \) is given by:
\[ \mu(x) = e^{\int p \, dx} = e^{\int -\frac{1}{x} \, dx} = e^{-\ln |x|} = \frac{1}{x} \]

### Step 4: Multiply through by the integrating factor
Multiplying the entire equation by the integrating factor:
\[ \frac{y}{x} - \frac{y}{x^2} = x^2 \cos x + x \sin x \]
This simplifies to:
\[ \frac{d}{dx}\left(\frac{y}{x}\right) = x \cos x + \sin x \]

### Step 5: Integrate both sides
Integrating both sides:
\[ \frac{y}{x} = \int (x \cos x + \sin x) \, dx \]

### Step 6: Solve the integral
Using integration by parts for \( \int x \cos x \, dx \):
Let \( u = x \) and \( dv = \cos x \, dx \):
\[ \int x \cos x \, dx = x \sin x - \int \sin x \, dx = x \sin x + \cos x \]
Thus,
\[ \int (x \cos x + \sin x) \, dx = x \sin x + \cos x - \cos x + C = x \sin x + C \]

### Step 7: Solve for \( y \)
So we have:
\[ \frac{y}{x} = x \sin x + C \]
Multiplying through by \( x \):
\[ y = x^2 \sin x + Cx \]

### Step 8: Apply the initial condition
Using the condition \( f(\pi) = \pi \):
\[ f(\pi) = \pi^2 \sin(\pi) + C\pi = \pi \]
Since \( \sin(\pi) = 0 \):
\[ C\pi = \pi \implies C = 1 \]
Thus, the function becomes:
\[ f(x) = x^2 \sin x + x \]

### Step 9: Find \( f(\frac{\pi}{2}) \)
Now, we calculate \( f(\frac{\pi}{2}) \):
\[ f\left(\frac{\pi}{2}\right) = \left(\frac{\pi}{2}\right)^2 \sin\left(\frac{\pi}{2}\right) + \frac{\pi}{2} = \frac{\pi^2}{4} + \frac{\pi}{2} \]

### Step 10: Find \( f''(x) \)
Next, we need to find \( f''(x) \):
1. First derivative:
\[ f'(x) = 2x \sin x + x^2 \cos x + 1 \]
2. Second derivative:
\[ f''(x) = 2 \sin x + 2x \cos x + 2x \cos x - x^2 \sin x = 2 \sin x + 4x \cos x - x^2 \sin x \]

### Step 11: Evaluate \( f''(\frac{\pi}{2}) \)
Now, we evaluate \( f''\left(\frac{\pi}{2}\right) \):
\[ f''\left(\frac{\pi}{2}\right) = 2 \sin\left(\frac{\pi}{2}\right) + 4\left(\frac{\pi}{2}\right) \cos\left(\frac{\pi}{2}\right) - \left(\frac{\pi}{2}\right)^2 \sin\left(\frac{\pi}{2}\right) \]
\[ = 2(1) + 0 - \frac{\pi^2}{4} = 2 - \frac{\pi^2}{4} \]

### Step 12: Calculate \( f''(\frac{\pi}{2}) + f(\frac{\pi}{2}) \)
Finally, we compute:
\[ f''\left(\frac{\pi}{2}\right) + f\left(\frac{\pi}{2}\right) = \left(2 - \frac{\pi^2}{4}\right) + \left(\frac{\pi^2}{4} + \frac{\pi}{2}\right) \]
\[ = 2 + \frac{\pi}{2} \]

Thus, the final answer is:
\[ \boxed{2 + \frac{\pi}{2}} \]