Q88 / 150 · MHT CET 2025 (25 Apr Shift 1) · Mathematics

$\mathrm{y}=\mathrm{e}^x(\mathrm{~A} \cos x+\mathrm{B} \sin x)$ is the solution of the differential equation

  • A
    $x^2 \frac{\mathrm{~d}^2 \mathrm{y}}{\mathrm{d} x^2}+\left(1+\mathrm{y}^2\right)=0$
  • B
    $\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{d} x^2}-\frac{\mathrm{dy}}{\mathrm{d} x}+\mathrm{y}=0$
  • Correct answer
    C
    $\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{d} x^2}-2 \frac{\mathrm{dy}}{\mathrm{d} x}+2 \mathrm{y}=0$
  • D
    $x \frac{\mathrm{~d}^2 \mathrm{y}}{\mathrm{d} x^2}-2 \frac{\mathrm{dy}}{\mathrm{d} x}+2 \mathrm{y}=0$

Solution

The differential equation for $y = e^x(A \cos x + B \sin x)$ can be found by differentiating twice to eliminate the arbitrary constants $A$ and $B$.

Differentiating $y$ with respect to $x$:

$\frac{dy}{dx} = e^x(A \cos x + B \sin x) + e^x(-A \sin x + B \cos x) = y + e^x(-A \sin x + B \cos x)$

Differentiating again:

$\frac{d^2y}{dx^2} = \frac{dy}{dx} + \frac{d}{dx}[e^x(-A \sin x + B \cos x)]$

Applying the product rule:

$\frac{d^2y}{dx^2} = \frac{dy}{dx} + e^x(-A \sin x + B \cos x) + e^x(-A \cos x - B \sin x)$

Substituting $e^x(-A \sin x + B \cos x) = \frac{dy}{dx} - y$ and $e^x(-A \cos x - B \sin x) = -y$:

$\frac{d^2y}{dx^2} = \frac{dy}{dx} + (\frac{dy}{dx} - y) - y = 2\frac{dy}{dx} - 2y$

The resulting differential equation is:

$\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y = 0$

Final answer: $\boxed{\text{C}}$

Hence, the correct answer is option: 'C'.

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