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If the solution curve of the differentiable equation dydx+2y=e3x\frac{dy}{dx} + 2y = e^{3x}, passes through the point (0,65)\left(0, \frac{6}{5}\right), then the value of y(log⁡e2)y(\log_e 2) is:

Solution

✅ Correct Option: 1

IF = e2xe^{2x}. Solution: y⋅e2x=∫e5xdx=e5x5+Cy \cdot e^{2x} = \int e^{5x} dx = \frac{e^{5x}}{5} + C.

y=e3x5+Ce−2xy = \frac{e^{3x}}{5} + Ce^{-2x}.

At x=0x=0: 65=15+C\frac{6}{5} = \frac{1}{5} + C, so C=1C = 1.

y(log⁡2)=85+14=32+520=3720y(\log 2) = \frac{8}{5} + \frac{1}{4} = \frac{32 + 5}{20} = \frac{37}{20}.

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