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Step-by-Step Solution
Step 1: Rewrite the given differential equation
We have the first-order linear differential equation:
$$x\,\frac{dy}{dx} - y = x^2\bigl(x \cos x + \sin x\bigr),\quad x>0.$$
We also have an initial condition given by:
$$y(\pi) = \pi.$$
Step 2: Convert to the standard linear form
Divide both sides of the differential equation by $x$, assuming $x \neq 0$:
$$\frac{dy}{dx} - \frac{1}{x}\,y = x\bigl(x \cos x + \sin x\bigr).$$
This is in the standard form:
$$\frac{dy}{dx} + P(x)\,y = Q(x),$$
where
$$P(x) = -\frac{1}{x}\quad\text{and}\quad Q(x) = x\bigl(x \cos x + \sin x\bigr).$$
Step 3: Find the integrating factor (I.F.)
The integrating factor for a linear first-order DE of the form
$$\frac{dy}{dx} + P(x)\,y = Q(x)$$
is given by
$$\text{I.F.} = e^{\int P(x)\,dx}.$$
Here,
$$P(x) = -\frac{1}{x},$$
so
$$\int P(x)\,dx = \int \left(-\frac{1}{x}\right) dx = -\ln x.$$
Thus, the integrating factor is
$$\text{I.F.} = e^{-\ln x} = \frac{1}{x}.$$
Step 4: Multiply through by the integrating factor
Multiply the original differential equation in its linear form by $\frac{1}{x}$:
$$\frac{1}{x}\,\frac{dy}{dx} - \frac{1}{x^2}\,y = \frac{1}{x}\,\Bigl[x\bigl(x\cos x + \sin x\bigr)\Bigr].$$
This simplifies to:
$$\frac{d}{dx}\Bigl(\frac{y}{x}\Bigr) = x \cos x + \sin x,$$
because multiplying by $\frac{1}{x}$ transforms the left side into the derivative of $\frac{y}{x}$.
Step 5: Integrate both sides to find $y$
We integrate with respect to $x$:
$$\frac{y}{x} = \int \bigl(x \cos x + \sin x\bigr)\,dx.$$
To integrate:
$$\int \left(x \cos x\right) dx,$$
we can use integration by parts, or recall a standard result. The integral of $x\cos x$ is $x \sin x + \cos x \cdot (-x)$ plus some terms, but let's do it systematically:
1. For $\int x \cos x\,dx$:
- Let $u = x$ and $dv = \cos x\,dx$.
- Then $du = dx$ and $v = \sin x$.
- By integration by parts: $\int u\,dv = uv - \int v\,du$, so
$$\int x \cos x\,dx = x \sin x - \int \sin x\,dx = x \sin x + \cos x + C_1.$$
2. For $\int \sin x\,dx = -\cos x + C_2.$
Hence,
$$\int \bigl(x \cos x + \sin x\bigr)dx = \bigl(x \sin x + \cos x\bigr) - \cos x + C = x \sin x + C,$$
where $C$ is the constant of integration (notice the $\cos x$ terms cancel out).
Therefore,
$$\frac{y}{x} = x \sin x + C \quad \Longrightarrow \quad y = x^2 \sin x + Cx.$$
Step 6: Apply the initial condition $y(\pi) = \pi$
We have $y(\pi) = \pi$, so substitute $x = \pi$:
$$y(\pi) = \pi^2 \sin(\pi) + C \pi.$$
Since $\sin(\pi) = 0$, this becomes:
$$\pi = 0 + C \pi \quad \Longrightarrow \quad C = 1.$$
Hence, the particular solution is
$$y = x^2 \sin x + x.$$
Step 7: Compute $y(\tfrac{\pi}{2})$
Substitute $x = \tfrac{\pi}{2}$,
$$y\Bigl(\tfrac{\pi}{2}\Bigr) = \Bigl(\tfrac{\pi}{2}\Bigr)^2 \sin\Bigl(\tfrac{\pi}{2}\Bigr) + \tfrac{\pi}{2}.$$
We know $\sin\bigl(\tfrac{\pi}{2}\bigr) = 1$, so
$$y\Bigl(\tfrac{\pi}{2}\Bigr) = \frac{\pi^2}{4} \cdot 1 + \frac{\pi}{2} = \frac{\pi^2}{4} + \frac{\pi}{2}.$$
Step 8: Find the second derivative $y''\bigl(\tfrac{\pi}{2}\bigr)$
First, let us find the first derivative $y'(x)$:
$$y(x) = x^2 \sin x + x.$$
Differentiate:
$$y'(x) = 2x \sin x + x^2 \cos x + 1.$$
Now differentiate again to get $y''(x)$:
\[
y''(x) = \frac{d}{dx}\bigl(2x\sin x + x^2 \cos x + 1\bigr).
\]
Compute term by term:
1. $\frac{d}{dx}\bigl(2x \sin x\bigr) = 2\sin x + 2x \cos x.$
2. $\frac{d}{dx}\bigl(x^2 \cos x\bigr)$:
- By product rule: $2x \cos x + x^2(-\sin x) = 2x\cos x - x^2\sin x.$
3. $\frac{d}{dx}(1) = 0.$
Hence,
$$y''(x) = 2\sin x + 2x \cos x + 2x \cos x - x^2 \sin x.$$
Simplify:
$$y''(x) = 2\sin x + 4x \cos x - x^2 \sin x.$$
Now evaluate at $x = \frac{\pi}{2}$:
\[
y''\Bigl(\tfrac{\pi}{2}\Bigr) = 2 \sin\Bigl(\tfrac{\pi}{2}\Bigr) + 4 \left(\tfrac{\pi}{2}\right)\cos\Bigl(\tfrac{\pi}{2}\Bigr)
- \left(\tfrac{\pi}{2}\right)^2 \sin\Bigl(\tfrac{\pi}{2}\Bigr).
\]
We have $\sin\bigl(\tfrac{\pi}{2}\bigr) = 1$ and $\cos\bigl(\tfrac{\pi}{2}\bigr) = 0$. So:
\[
y''\Bigl(\tfrac{\pi}{2}\Bigr) = 2 \cdot 1 + 4\left(\tfrac{\pi}{2}\right) \cdot 0 - \left(\tfrac{\pi}{2}\right)^2 \cdot 1
= 2 - \frac{\pi^2}{4}.
\]
Step 9: Evaluate the quantity $y\bigl(\tfrac{\pi}{2}\bigr) + y''\bigl(\tfrac{\pi}{2}\bigr)$
From the previous results:
\[
y\Bigl(\tfrac{\pi}{2}\Bigr) = \frac{\pi^2}{4} + \frac{\pi}{2},
\quad\text{and}\quad
y''\Bigl(\tfrac{\pi}{2}\Bigr) = 2 - \frac{\pi^2}{4}.
\]
Adding these,
\[
y\Bigl(\tfrac{\pi}{2}\Bigr) + y''\Bigl(\tfrac{\pi}{2}\Bigr)
= \left(\frac{\pi^2}{4} + \frac{\pi}{2}\right) + \left(2 - \frac{\pi^2}{4}\right).
\]
Notice that $ \frac{\pi^2}{4}$ and $ - \frac{\pi^2}{4}$ cancel. Hence,
\[
y\Bigl(\tfrac{\pi}{2}\Bigr) + y''\Bigl(\tfrac{\pi}{2}\Bigr) = 2 + \frac{\pi}{2}.
\]
Final Answer
$$y''\!\Bigl(\tfrac{\pi}{2}\Bigr) + y\!\Bigl(\tfrac{\pi}{2}\Bigr) = 2 + \frac{\pi}{2}.$$
