For solving ` dy/dx = 4x +y +1`, suitable substitution is

Options

A. y = vx
B. y = 4x
C. y = 4x + v
D. y + 4x + 1 = v

Answer

To solve the differential equation \( \frac{dy}{dx} = 4x + y + 1 \), we will use a suitable substitution. Let's go through the steps systematically.

### Step 1: Identify a suitable substitution
We observe that the right-hand side of the equation contains both \( y \) and \( x \). A good substitution to simplify this equation is to let:
\[ v = 4x + y + 1 \]

### Step 2: Differentiate the substitution
Next, we differentiate \( v \) with respect to \( x \):
\[ \frac{dv}{dx} = \frac{d}{dx}(4x + y + 1) \]
Using the chain rule, we have:
\[ \frac{dv}{dx} = 4 + \frac{dy}{dx} \]

### Step 3: Substitute \( \frac{dy}{dx} \) in the original equation
From the original equation, we can express \( \frac{dy}{dx} \) in terms of \( v \):
\[ \frac{dy}{dx} = \frac{dv}{dx} - 4 \]

### Step 4: Substitute into the original equation
Now we substitute \( \frac{dy}{dx} \) back into the original equation:
\[ \frac{dv}{dx} - 4 = 4x + y + 1 \]
Since we defined \( v = 4x + y + 1 \), we can replace \( 4x + y + 1 \) with \( v \):
\[ \frac{dv}{dx} - 4 = v \]

### Step 5: Rearrange the equation
Rearranging gives us:
\[ \frac{dv}{dx} = v + 4 \]

### Step 6: Separate the variables
Now we separate the variables:
\[ \frac{dv}{v + 4} = dx \]

### Step 7: Integrate both sides
Integrating both sides, we have:
\[ \int \frac{1}{v + 4} dv = \int dx \]
This results in:
\[ \ln |v + 4| = x + C \]
where \( C \) is the constant of integration.

### Step 8: Solve for \( v \)
Exponentiating both sides gives:
\[ |v + 4| = e^{x + C} = e^C e^x \]
Let \( K = e^C \), then:
\[ v + 4 = K e^x \]

### Step 9: Substitute back for \( y \)
Recalling our substitution \( v = 4x + y + 1 \), we substitute back:
\[ 4x + y + 1 = K e^x \]
Thus,
\[ y = K e^x - 4x - 1 \]

### Conclusion
The solution to the differential equation \( \frac{dy}{dx} = 4x + y + 1 \) is:
\[ y = K e^x - 4x - 1 \]