CUET UG 2025 — Mathematics Calculus

Given $y = 3e^{2x} + 2e^{3x}$y=3e2x+2e3x, find $\frac{d^2y}{dx^2} + 6y$dx2d2y​+6y.

When differentiating $e^{ax}$eax, the derivative is $a \cdot e^{ax}$a⋅eax.

$$\frac{dy}{dx} = 3 \cdot (2e^{2x}) + 2 \cdot (3e^{3x})$$dxdy​=3⋅(2e2x)+2⋅(3e3x)

$$\frac{dy}{dx} = 6e^{2x} + 6e^{3x}$$dxdy​=6e2x+6e3x

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Find the second derivative:

$$\frac{d^2y}{dx^2} = 6 \cdot (2e^{2x}) + 6 \cdot (3e^{3x})$$dx2d2y​=6⋅(2e2x)+6⋅(3e3x)

$$\frac{d^2y}{dx^2} = 12e^{2x} + 18e^{3x}$$dx2d2y​=12e2x+18e3x

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Calculate $\frac{d^2y}{dx^2} + 6y$dx2d2y​+6y:

$$\frac{d^2y}{dx^2} + 6y = (12e^{2x} + 18e^{3x}) + 6(3e^{2x} + 2e^{3x})$$dx2d2y​+6y=(12e2x+18e3x)+6(3e2x+2e3x)

$$= 12e^{2x} + 18e^{3x} + 18e^{2x} + 12e^{3x}$$=12e2x+18e3x+18e2x+12e3x

$$= 30e^{2x} + 30e^{3x}$$=30e2x+30e3x

$$= 30(e^{2x} + e^{3x})$$=30(e2x+e3x)

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From the first derivative, $\frac{dy}{dx} = 6e^{2x} + 6e^{3x} = 6(e^{2x} + e^{3x})$dxdy​=6e2x+6e3x=6(e2x+e3x)

Therefore:

$$30(e^{2x} + e^{3x}) = 5 \times 6(e^{2x} + e^{3x})$$30(e2x+e3x)=5×6(e2x+e3x)

$$= 5\frac{dy}{dx}$$=5dxdy​

Therefore, $\frac{d^2y}{dx^2} + 6y = 5\frac{dy}{dx}$dx2d2y​+6y=5dxdy​