CUET UG 2025 — Mathematics Calculus

Given $\log_e\left(\frac{dy}{dx}\right) = 3x + 4y$loge​(dxdy​)=3x+4y

If $\log_e(A) = B$loge​(A)=B, then $A = e^B$A=eB

$$\frac{dy}{dx} = e^{3x + 4y}$$dxdy​=e3x+4y

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

$$\frac{dy}{dx} = e^{3x} \cdot e^{4y}$$dxdy​=e3x⋅e4y

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Separating the variables:

$$\frac{1}{e^{4y}} \, dy = e^{3x} \, dx$$e4y1​dy=e3xdx

$$e^{-4y} \, dy = e^{3x} \, dx$$e−4ydy=e3xdx

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Integrating both sides:

$$\int e^{-4y} \, dy = \int e^{3x} \, dx$$∫e−4ydy=∫e3xdx

$$-\frac{e^{-4y}}{4} = \frac{e^{3x}}{3} + C_1$$−4e−4y​=3e3x​+C1​

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Multiplying by $-12$−12 to clear fractions:

$$3e^{-4y} = -4e^{3x} - 12C_1$$3e−4y=−4e3x−12C1​

$$4e^{3x} + 3e^{-4y} = -12C_1$$4e3x+3e−4y=−12C1​

Since $C_1$C1​ is arbitrary, let $C = -12C_1$C=−12C1​:

$$4e^{3x} + 3e^{-4y} = C$$4e3x+3e−4y=C

Therefore, the solution is $4e^{3x} + 3e^{-4y} + C = 0$4e3x+3e−4y+C=0