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: Conceptualization:
This is a first-order differential equation solvable via separation of variables.
Step 2: Methodology:
The technique involves isolating terms with 'y' and 'dy' on one side and terms with 'x' and 'dx' on the other, followed by integration of both sides.
Step 3: Procedural Breakdown:
The differential equation is:
\[ \log_e\left(\frac{dy}{dx}\right) = 3x + 4y \] Exponentiate both sides to eliminate the logarithm:
\[ \frac{dy}{dx} = e^{3x + 4y} \] Apply the exponent property \(e^{a+b} = e^a \cdot e^b\):
\[ \frac{dy}{dx} = e^{3x} \cdot e^{4y} \] Separate variables:
\[ \frac{dy}{e^{4y}} = e^{3x} dx \] \[ e^{-4y} dy = e^{3x} dx \] Integrate both sides:
\[ \int e^{-4y} dy = \int e^{3x} dx \] Perform integration:
\[ \frac{e^{-4y}}{-4} = \frac{e^{3x}}{3} + C_1 \] where \(C_1\) is the constant of integration. Rearrange to match option formats. Move the x-term to the left:
\[ -\frac{e^{-4y}}{4} - \frac{e^{3x}}{3} = C_1 \] Multiply by -12 to clear fractions:
\[ (-12) \left(-\frac{e^{-4y}}{4}\right) - (-12) \left(\frac{e^{3x}}{3}\right) = -12 C_1 \] \[ 3e^{-4y} + 4e^{3x} = -12 C_1 \] Substitute \(C = -12 C_1\), where C is an arbitrary constant:
\[ 4e^{3x} + 3e^{-4y} = C \] This can be expressed as:
\[ 4e^{3x} + 3e^{-4y} - C = 0 \] Replace -C with a new arbitrary constant, also denoted as C:
\[ 4e^{3x} + 3e^{-4y} + C = 0 \] Step 4: Conclusion:
The solved differential equation is $4e^{3x + 3e^{-4y} + C = 0$}.