Differential Equation P.I. Calculation

We are given the differential equation $(D^2-2D+4)y = e^x \sin x$. The task is to find the particular integral (P.I.) which is given in the form $\frac{1}{2}e^x f(x)$, and then determine the characteristics of the function $f(x)$.

Finding the Particular Integral (P.I.)

We can calculate the P.I. using the operator method. The equation is:

P.I. = $\frac{1}{D^2-2D+4} (e^x \sin x)$

To handle the $e^x$ term, we use the shifting property of the operator: $\frac{1}{F(D)} (e^{ax} \phi(x)) = e^{ax} \frac{1}{F(D+a)} \phi(x)$.

Here, $a=1$ and $\phi(x) = \sin x$. The operator $F(D)$ is $D^2-2D+4$. We need to calculate $F(D+a) = F(D+1)$:

$F(D+1) = (D+1)^2 - 2(D+1) + 4$

Expanding this expression gives:

$F(D+1) = (D^2 + 2D + 1) - (2D + 2) + 4$

$F(D+1) = D^2 + 2D + 1 - 2D - 2 + 4$

$F(D+1) = D^2 + 3$

Now, we substitute this back into the P.I. expression:

P.I. = $e^x \frac{1}{D^2+3} (\sin x)$

To evaluate the remaining part, $\frac{1}{D^2+3} (\sin x)$, we use the rule for integrating $\sin(kx)$ or $\cos(kx)$, which requires replacing $D^2$ with $-k^2$. In this case, $k=1$ because we have $\sin(1x)$.

Substitute $D^2 = -(1)^2 = -1$ into the denominator:

$\frac{1}{D^2+3} (\sin x) = \frac{1}{-1+3} \sin x = \frac{1}{2} \sin x$.

This substitution is valid because the denominator ($D^2+3$) does not become zero when $D^2$ is replaced by $-1$. (-1 + 3 = 2).

Therefore, the P.I. is:

P.I. = $e^x \left( \frac{1}{2} \sin x \right) = \frac{1}{2} e^x \sin x$

Function $f(x)$ Identification

The problem states that the P.I. is given in the form $\frac{1}{2}e^x f(x)$.

By comparing our calculated P.I., $\frac{1}{2} e^x \sin x$, with the given form $\frac{1}{2} e^x f(x)$, we can directly identify $f(x)$:

$f(x) = \sin x$

Function $f(x)$ Property Analysis

We now need to analyze the properties of $f(x) = \sin x$ on the interval $[0, \pi]$ as described in the options.

First, let's find the derivative of $f(x)$ to understand its behavior:

$f'(x) = \frac{d}{dx}(\sin x) = \cos x$

Now, let's examine the sign of $f'(x)$ on the interval $[0, \pi]$:

• For $x$ in the interval $[0, \frac{\pi}{2})$, $f'(x) = \cos x$ is positive ($f'(x) > 0$). This means $f(x) = \sin x$ is increasing on this sub-interval.
• At $x = \frac{\pi}{2}$, $f'(x) = \cos(\frac{\pi}{2}) = 0$.
• For $x$ in the interval $(\frac{\pi}{2}, \pi]$, $f'(x) = \cos x$ is negative ($f'(x) < 0$). This means $f(x) = \sin x$ is decreasing on this sub-interval.

Based on this analysis, let's evaluate the given options:

• Option 1: an increasing function on $[0, \pi]$
  This statement is incorrect because $f(x) = \sin x$ decreases on the interval $(\frac{\pi}{2}, \pi]$.
• Option 2: a decreasing function on $[0, \pi]$
  This statement is incorrect because $f(x) = \sin x$ increases on the interval $[0, \frac{\pi}{2})$.
• Option 3: a continuous function on $[-2\pi, 2\pi]$
  The function $f(x) = \sin x$ is continuous everywhere on the real number line. Therefore, it is continuous on the interval $[-2\pi, 2\pi]$. This statement is correct.
• Option 4: not differentiable function at $x = 0$
  The function $f(x) = \sin x$ is differentiable everywhere. Its derivative at $x=0$ is $f'(0) = \cos(0) = 1$. Thus, this statement is incorrect.