Question:medium

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

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$\log(A) = B \iff A = e^B$.
Updated On: May 14, 2026
  • $2e^{2x} + 5e^{5y} = 6$
  • $5e^{2x} - 2e^{5y} = 3$
  • $2e^{2x} - 5e^{5y} = 6$
  • $5e^{2x} + 2e^{5y} = 3$
Show Solution

The Correct Option is D

Solution and Explanation

Step 1: Understanding the Question:
We need to find the particular solution of a first-order separable differential equation given an initial condition.
Step 3: Detailed Explanation:
1. Start with the given equation: $\log(\frac{dy}{dx}) = 2x - 5y$.
Take exponent on both sides: $\frac{dy}{dx} = e^{2x - 5y} = e^{2x} \cdot e^{-5y}$.
2. Separate the variables:
$\frac{1}{e^{-5y}} dy = e^{2x} dx \implies e^{5y} dy = e^{2x} dx$.
3. Integrate both sides:
$\int e^{5y} dy = \int e^{2x} dx$
$\frac{e^{5y}}{5} = \frac{e^{2x}}{2} + c \implies 2e^{5y} = 5e^{2x} + 10c$.
Let $C = -10c$ be a new constant: $5e^{2x} - 2e^{5y} = C$.
4. Use initial condition $y(0) = 0$:
$5e^{2(0)} - 2e^{5(0)} = C \implies 5(1) - 2(1) = C \implies C = 3$.
5. Final equation: $5e^{2x} - 2e^{5y} = 3$.
Step 4: Final Answer:
The solution is $5e^{2x} - 2e^{5y} = 3$.
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