Question

Solve the following differential equations.

$\ln \frac{dy}{dx} = ax + by$

• A. $ae^{-by} + be^{ax} = C$
• B. $ae^{by} + be^{-ax} = C$
• C. $be^{by} - ae^{ax} = C$
• D. None of these

Answer 1

Answer: (A)

A $ae^{-by} + be^{ax} = C$

Answer 2

The given differential equation is: $\ln \frac{dy}{dx} = ax + by$

Step 1: Convert the logarithmic equation into an exponential form. To remove the natural logarithm, we exponentiate both sides with base $e$: $e^{\ln \frac{dy}{dx}} = e^{ax + by}$ $\frac{dy}{dx} = e^{ax + by}$

Step 2: Separate the variables. Using the property of exponents $e^{m+n} = e^m \cdot e^n$, we can write: $\frac{dy}{dx} = e^{ax} \cdot e^{by}$

Now, separate the variables $x$ and $y$ by moving all terms involving $y$ to one side with $dy$ and all terms involving $x$ to the other side with $dx$: $\frac{dy}{e^{by}} = e^{ax} dx$ We can rewrite $\frac{1}{e^{by}}$ as $e^{-by}$: $e^{-by} dy = e^{ax} dx$

Step 3: Integrate both sides. Now, integrate both sides of the equation: $\int e^{-by} dy = \int e^{ax} dx$

For the left side, $\int e^{-by} dy$: The integral of $e^{kx}$ is $\frac{1}{k}e^{kx}$. Here $k = -b$. $\int e^{-by} dy = \frac{e^{-by}}{-b} + C_1 = -\frac{1}{b} e^{-by} + C_1$

For the right side, $\int e^{ax} dx$: Here $k = a$. $\int e^{ax} dx = \frac{e^{ax}}{a} + C_2$

Equating the results of the integration: $-\frac{1}{b} e^{-by} = \frac{1}{a} e^{ax} + C'$ (where $C'$ is the combined constant of integration, $C' = C_2 - C_1$)

Step 4: Rearrange the terms to match the given options. To eliminate the denominators $a$ and $b$, multiply the entire equation by $ab$: $ab \left(-\frac{1}{b} e^{-by}\right) = ab \left(\frac{1}{a} e^{ax} + C'\right)$ $-a e^{-by} = b e^{ax} + ab C'$

Now, rearrange the terms to match the format of the options. Move all terms involving $e^{ax}$ and $e^{-by}$ to one side: $0 = b e^{ax} + a e^{-by} + ab C'$

Let $C = -ab C'$. Since $C'$ is an arbitrary constant, $C$ is also an arbitrary constant. So, the solution becomes: $b e^{ax} + a e^{-by} = C$

This matches option A.