Question

The solution of $\log(\frac{dy}{dx}) = 2x - 5y, y(0) = 0$ is

• A. $2e^{2x} + 5e^{5y} = 6$
• B. $5e^{2x} - 2e^{5y} = 3$
• C. $2e^{2x} - 5e^{5y} = 6$
• D. $5e^{2x} + 2e^{5y} = 3$

Hint:

$\log(A) = B \iff A = e^B$.

Answer 1

The Correct Option is D

Step 1: Remove Logarithm
$\frac{dy}{dx} = e^{2x - 5y} = e^{2x} \cdot e^{-5y}$.

Step 2: Separate Variables
$e^{5y} dy = e^{2x} dx$.

Step 3: Integrate
$\int e^{5y} dy = \int e^{2x} dx \implies \frac{e^{5y}}{5} = \frac{e^{2x}}{2} + C$.
Multiplying by 10: $2e^{5y} = 5e^{2x} + 10C$.

Step 4: Use Initial Condition
$y(0) = 0 \implies 2e^{0} = 5e^{0} + K \implies 2 = 5 + K \implies K = -3$.
$2e^{5y} = 5e^{2x} - 3 \implies 5e^{2x} - 2e^{5y} = 3$.
Final Answer: (B)