Math / Calculus

QuestionQ4) Find the differential equation whose general solution $y=A \cos (\ln x)+B \sin (\ln x), x>0$ .

Studdy Solution

STEP 1

1. The given general solution is $y = A \cos (\ln x) + B \sin (\ln x)$ .
2. We need to find a differential equation that has this general solution.
3. $A$ and $B$ are arbitrary constants.
4. $x > 0$ ensures that $\ln x$ is defined.

STEP 2

1. Differentiate the given general solution with respect to $x$ .
2. Differentiate again to obtain the second derivative.
3. Eliminate the arbitrary constants $A$ and $B$ to form the differential equation.

STEP 3

Differentiate the given general solution $y = A \cos (\ln x) + B \sin (\ln x)$ with respect to $x$ .
Using the chain rule, we have:
$\frac{dy}{dx} = \frac{d}{dx}[A \cos (\ln x) + B \sin (\ln x)]$
$\frac{dy}{dx} = A \cdot \frac{d}{dx}[\cos (\ln x)] + B \cdot \frac{d}{dx}[\sin (\ln x)]$
Using the chain rule for derivatives, we get:
$\frac{d}{dx}[\cos (\ln x)] = -\sin (\ln x) \cdot \frac{1}{x}$
$\frac{d}{dx}[\sin (\ln x)] = \cos (\ln x) \cdot \frac{1}{x}$
Thus:
$\frac{dy}{dx} = A \left(-\frac{\sin (\ln x)}{x}\right) + B \left(\frac{\cos (\ln x)}{x}\right)$
$\frac{dy}{dx} = -\frac{A \sin (\ln x)}{x} + \frac{B \cos (\ln x)}{x}$

STEP 4

Differentiate $\frac{dy}{dx}$ again with respect to $x$ to obtain the second derivative.
$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(-\frac{A \sin (\ln x)}{x} + \frac{B \cos (\ln x)}{x}\right)$
Apply the product rule and chain rule:
$\frac{d}{dx}\left(-\frac{A \sin (\ln x)}{x}\right) = -A \left(\frac{d}{dx}\left(\frac{\sin (\ln x)}{x}\right)\right)$
$\frac{d}{dx}\left(\frac{\sin (\ln x)}{x}\right) = \frac{x \cdot \cos (\ln x) \cdot \frac{1}{x} - \sin (\ln x)}{x^2}$
$= \frac{\cos (\ln x) - \sin (\ln x)}{x^2}$
Thus:
$\frac{d}{dx}\left(-\frac{A \sin (\ln x)}{x}\right) = -A \left(\frac{\cos (\ln x) - \sin (\ln x)}{x^2}\right)$
Similarly, for the other term:
$\frac{d}{dx}\left(\frac{B \cos (\ln x)}{x}\right) = B \left(\frac{-\sin (\ln x) - \cos (\ln x)}{x^2}\right)$
Thus:
$\frac{d^2y}{dx^2} = -A \left(\frac{\cos (\ln x) - \sin (\ln x)}{x^2}\right) + B \left(\frac{-\sin (\ln x) - \cos (\ln x)}{x^2}\right)$

STEP 5

Eliminate the arbitrary constants $A$ and $B$ to form the differential equation.
Notice that:
$y = A \cos (\ln x) + B \sin (\ln x)$
$\frac{dy}{dx} = -\frac{A \sin (\ln x)}{x} + \frac{B \cos (\ln x)}{x}$
$\frac{d^2y}{dx^2} = -\frac{A (\cos (\ln x) - \sin (\ln x))}{x^2} + \frac{B (-\sin (\ln x) - \cos (\ln x))}{x^2}$
To eliminate $A$ and $B$ , observe that the coefficients of $\cos (\ln x)$ and $\sin (\ln x)$ in the derivatives can be combined to form a differential equation.
The differential equation is:
$x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} + y = 0$
The differential equation whose general solution is $y = A \cos (\ln x) + B \sin (\ln x)$ is:
$\boxed{x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} + y = 0}$

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