Question:
The general solution for the differential equation \( \frac{dy}{dx} = e^{3x - y} \) is :
The general solution for the differential equation \( \frac{dy}{dx} = e^{3x - y} \) is :
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When exponents contain sums or differences like \( e^{a-b} \), separating them into products or quotients \( \frac{e^a}{e^b} \) is the standard first step to isolate variables.
Updated On: Jul 6, 2026
- \( 3e^y = e^{3x} + C \)
- \( \log(3x - y) = C \)
- \( e^{3x - y} = C \)
- \( -e^y + 3e^{3x} = C \)
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The Correct Option is A
Solution and Explanation
Concept:
This differential equation can be solved using the method of separation of variables. We transform the equation into the form \( g(y) \, dy = f(x) \, dx \), and then integrate both sides independently.
Step 1: Separate the variables \( x \) and \( y \). The given differential equation is: \[ \frac{dy}{dx} = e^{3x - y} \] Using exponential laws, we split the right-hand side: \[ \frac{dy}{dx} = \frac{e^{3x}}{e^y} \] Now, cross-multiply to separate variables on opposite sides: \[ e^y \, dy = e^{3x} \, dx \]
Step 2: Integrate both sides. Set up the integrals: \[ \int e^y \, dy = \int e^{3x} \, dx \] Evaluating the basic exponential integrals yields: \[ e^y = \frac{e^{3x}}{3} + C_1 \]
Step 3: Clear fractions to match option formatting. Multiply the entire equation by \( 3 \): \[ 3e^y = e^{3x} + 3C_1 \] Let \( 3C_1 = C \) be a new arbitrary constant: \[ 3e^y = e^{3x} + C \]
Step 1: Separate the variables \( x \) and \( y \). The given differential equation is: \[ \frac{dy}{dx} = e^{3x - y} \] Using exponential laws, we split the right-hand side: \[ \frac{dy}{dx} = \frac{e^{3x}}{e^y} \] Now, cross-multiply to separate variables on opposite sides: \[ e^y \, dy = e^{3x} \, dx \]
Step 2: Integrate both sides. Set up the integrals: \[ \int e^y \, dy = \int e^{3x} \, dx \] Evaluating the basic exponential integrals yields: \[ e^y = \frac{e^{3x}}{3} + C_1 \]
Step 3: Clear fractions to match option formatting. Multiply the entire equation by \( 3 \): \[ 3e^y = e^{3x} + 3C_1 \] Let \( 3C_1 = C \) be a new arbitrary constant: \[ 3e^y = e^{3x} + C \]
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