CUET UG 2025 — Mathematics Calculus

Given:

$$\log_e\left(\frac{dy}{dx}\right) = 5x + 2y$$loge​(dxdy​)=5x+2y

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Apply $e$e to both sides:

$$e^{\log_e\left(\frac{dy}{dx}\right)} = e^{5x + 2y}$$eloge​(dxdy​)=e5x+2y

$$\frac{dy}{dx} = e^{5x + 2y}$$dxdy​=e5x+2y

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Using $e^{a+b} = e^a \cdot e^b$ea+b=ea⋅eb:

$$\frac{dy}{dx} = e^{5x} \cdot e^{2y}$$dxdy​=e5x⋅e2y

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Separating variables:

$$\frac{dy}{e^{2y}} = e^{5x} \, dx$$e2ydy​=e5xdx

$$e^{-2y} \, dy = e^{5x} \, dx$$e−2ydy=e5xdx

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Integrating both sides:

$$\int e^{-2y} \, dy = \int e^{5x} \, dx$$∫e−2ydy=∫e5xdx

$$-\frac{1}{2}e^{-2y} = \frac{e^{5x}}{5} + C_1$$−21​e−2y=5e5x​+C1​

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Multiplying both sides by $-10$−10:

$$5e^{-2y} = -2e^{5x} - 10C_1$$5e−2y=−2e5x−10C1​

$$2e^{5x} + 5e^{-2y} = -10C_1$$2e5x+5e−2y=−10C1​

Since $-10C_1$−10C1​ is an arbitrary constant, let $C = -10C_1$C=−10C1​:

$$2e^{5x} + 5e^{-2y} + C = 0$$2e5x+5e−2y+C=0