Question

If the integrating factor of $\frac{dy}{dx} + \left[(x^2 - 2x)\cos x + 2(x - 1)\sin x\right]y = x^2$ is $e^{f(x)}$, then $f(3) =$

• A. $3\cos 3 + 4\sin 3$
• B. $4\cos 3$
• C. $3\sin 3$
• D. $4\sin 3 + \cos 3$

Hint:

Always look for exact differential patterns like $d(u \cdot v) = u \, dv + v \, du$ when dealing with long, intimidating integrals involving combinations of polynomials and trigonometric functions.

Answer 1

The Correct Option is C

Concept: The given differential equation is a first-order linear differential equation of the form: \[ \frac{dy}{dx} + P(x)y = Q(x) \] where $P(x) = (x^2 - 2x)\cos x + 2(x - 1)\sin x$. The integrating factor ($I.F.$) of such an equation is given by: \[ I.F. = e^{\int P(x) \, dx} \] We are given that $I.F. = e^{f(x)}$, which implies: \[ f(x) = \int P(x) \, dx = \int \left[ (x^2 - 2x)\cos x + 2(x - 1)\sin x \right] dx \]

Step 1: Evaluate the integral using integration by parts or recognizing an exact derivative structure.
Let us look closely at the terms inside the integrand: Notice that the derivative of $(x^2 - 2x)$ is $2(x - 1)$, and the integral of $\cos x$ is $\sin x$. This strongly suggests the application of the product rule for differentiation, namely $\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$. Let us test the derivative of the function $g(x) = (x^2 - 2x)\sin x$: \[ \frac{d}{dx}\left[ (x^2 - 2x)\sin x \right] = \frac{d}{dx}(x^2 - 2x) \cdot \sin x + (x^2 - 2x) \cdot \frac{d}{dx}(\sin x) \] \[ = 2(x - 1)\sin x + (x^2 - 2x)\cos x \] This expression matches the integrand perfectly!

Step 2: Write down the expression for $f(x)$.
Since the integrand is the exact derivative of $(x^2 - 2x)\sin x$, the integral simplifies to: \[ f(x) = (x^2 - 2x)\sin x \] (We omit the constant of integration as it is conventional when determining specific standard forms of functions for evaluating values).

Step 3: Calculate $f(3)$.
Substitute $x = 3$ into the function $f(x)$: \[ f(3) = (3^2 - 2(3))\sin 3 \] \[ f(3) = (9 - 6)\sin 3 = 3\sin 3 \]