Question

The solution of the differential equation $\log_e(\frac{dy}{dx}) = 3x + 4y$ is given by

• A. $4e^{3x} + 3e^{-4y} + C = 0$, where C is constant of integration
• B. $3e^{3x} + 4e^{-4y} + C = 0$, where C is constant of integration
• C. $4e^{-3x} + 3e^{4y} + C = 0$, where C is constant of integration
• D. $3e^{-3x} + 4e^{4y} + C = 0$, where C is constant of integration

Hint:

When solving differential equations, remember that the constant of integration C is arbitrary. An expression like `... = C` is equivalent to `... - C = 0` or `... + C = 0`, as the sign and magnitude of an arbitrary constant are also arbitrary. Match your final form to the given options.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This is a first-order differential equation that can be solved using the method of separation of variables.
Step 2: Key Formula or Approach:
The method involves rearranging the equation so that all terms involving y and dy are on one side, and all terms involving x and dx are on the other. Then, integrate both sides.
Step 3: Detailed Explanation:
The given differential equation is:
\[ \log_e\left(\frac{dy}{dx}\right) = 3x + 4y \] To remove the logarithm, we take the exponential of both sides:
\[ \frac{dy}{dx} = e^{3x + 4y} \] Using the property of exponents \(e^{a+b} = e^a \cdot e^b\), we can write:
\[ \frac{dy}{dx} = e^{3x} \cdot e^{4y} \] Now, we separate the variables by moving all y-terms to the left side and all x-terms to the right side:
\[ \frac{dy}{e^{4y}} = e^{3x} dx \] \[ e^{-4y} dy = e^{3x} dx \] Integrate both sides of the equation:
\[ \int e^{-4y} dy = \int e^{3x} dx \] Performing the integration:
\[ \frac{e^{-4y}}{-4} = \frac{e^{3x}}{3} + C_1 \] where \(C_1\) is the constant of integration. To match the form of the options, we can rearrange the terms. Move the x-term to the left side:
\[ -\frac{e^{-4y}}{4} - \frac{e^{3x}}{3} = C_1 \] Multiply the entire equation by -12 to eliminate the fractions:
\[ (-12) \left(-\frac{e^{-4y}}{4}\right) - (-12) \left(\frac{e^{3x}}{3}\right) = -12 C_1 \] \[ 3e^{-4y} + 4e^{3x} = -12 C_1 \] Let \(C = -12 C_1\). Since \(C_1\) is an arbitrary constant, C is also an arbitrary constant.
\[ 4e^{3x} + 3e^{-4y} = C \] This can be written as:
\[ 4e^{3x} + 3e^{-4y} - C = 0 \] Since C is an arbitrary constant, we can replace -C with a new constant, which we can also call C, giving the final form:
\[ 4e^{3x} + 3e^{-4y} + C = 0 \] Step 4: Final Answer:
The solution to the differential equation is $4e^{3x + 3e^{-4y} + C = 0$}.