Find the particular solution of :
(i) `log(dy/dx)=2x+y` (ii) `log(dy/dx)=ax+by`, given that `y=0` when `x=0`.

Answer

To find the particular solutions for the given differential equations, we will solve each equation step by step.

### Part (i): Solve `log(dy/dx) = 2x + y`

1. **Exponentiate both sides to eliminate the logarithm:**
\[ \frac{dy}{dx} = e^{2x + y} \]

2. **Rewrite the equation:**
\[ \frac{dy}{dx} = e^{2x} \cdot e^{y} \]

3. **Separate the variables:**
\[ e^{-y} dy = e^{2x} dx \]

4. **Integrate both sides:**
\[ \int e^{-y} dy = \int e^{2x} dx \]
The left side integrates to:
\[ -e^{-y} = \frac{1}{2} e^{2x} + C \]

5. **Rearranging gives:**
\[ e^{-y} = -\frac{1}{2} e^{2x} - C \]

6. **Taking the reciprocal:**
\[ e^{y} = -\frac{1}{\frac{1}{2} e^{2x} + C} \]

7. **To find the particular solution, use the initial condition \(y(0) = 0\):**
\[ e^{0} = -\frac{1}{\frac{1}{2} e^{0} + C} \]
This simplifies to:
\[ 1 = -\frac{1}{\frac{1}{2} + C} \]
Solving for \(C\):
\[ \frac{1}{2} + C = -1 \implies C = -\frac{3}{2} \]

8. **Substituting \(C\) back into the equation gives the particular solution:**
\[ e^{y} = -\frac{1}{\frac{1}{2} e^{2x} - \frac{3}{2}} \]

### Part (ii): Solve `log(dy/dx) = ax + by`

1. **Exponentiate both sides:**
\[ \frac{dy}{dx} = e^{ax + by} \]

2. **Rewrite the equation:**
\[ \frac{dy}{dx} = e^{ax} \cdot e^{by} \]

3. **Separate the variables:**
\[ e^{-by} dy = e^{ax} dx \]

4. **Integrate both sides:**
\[ \int e^{-by} dy = \int e^{ax} dx \]
The left side integrates to:
\[ -\frac{1}{b} e^{-by} = \frac{1}{a} e^{ax} + C \]

5. **Rearranging gives:**
\[ e^{-by} = -\frac{b}{a} e^{ax} - bC \]

6. **Taking the reciprocal:**
\[ e^{by} = -\frac{1}{-\frac{b}{a} e^{ax} - bC} \]

7. **To find the particular solution, use the initial condition \(y(0) = 0\):**
\[ e^{0} = -\frac{1}{-\frac{b}{a} e^{0} - bC} \]
This simplifies to:
\[ 1 = -\frac{1}{-\frac{b}{a} - bC} \]
Solving for \(C\):
\[ -\frac{b}{a} - bC = -1 \implies bC = -\frac{b}{a} + 1 \implies C = \frac{1 - \frac{b}{a}}{b} \]

8. **Substituting \(C\) back into the equation gives the particular solution:**
\[ e^{by} = -\frac{1}{-\frac{b}{a} e^{ax} + \left(1 - \frac{b}{a}\right)} \]