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

Options

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

Answer

To solve the differential equation \( \frac{dy}{dx} = 4x + y + 1 \), we can use a suitable substitution. Here are the steps to find the substitution and solve the equation:

### Step 1: Identify the substitution
We can let:
\[ b = 4x + y + 1 \]
This substitution simplifies the equation by combining the variables.

### Step 2: Differentiate the substitution
Next, we differentiate both sides with respect to \( x \):
\[ \frac{db}{dx} = 4 + \frac{dy}{dx} \]
Here, the derivative of \( 4x \) is \( 4 \), and the derivative of \( y \) with respect to \( x \) is \( \frac{dy}{dx} \).

### Step 3: Rearrange the equation
Now, we can rearrange the equation to express \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{db}{dx} - 4 \]

### Step 4: Substitute back into the original equation
Now, substitute \( b \) back into the original differential equation:
\[ \frac{db}{dx} - 4 = b \]
This simplifies to:
\[ \frac{db}{dx} = b + 4 \]

### Step 5: Separate variables
We can rewrite this as:
\[ \frac{1}{b + 4} db = dx \]

### Step 6: Integrate both sides
Now, we integrate both sides:
\[ \int \frac{1}{b + 4} db = \int dx \]
This gives us:
\[ \ln |b + 4| = x + C \]
where \( C \) is the constant of integration.

### Step 7: Exponentiate to solve for \( b \)
Exponentiating both sides, we get:
\[ b + 4 = e^{x + C} = e^C e^x \]
Let \( K = e^C \), then:
\[ b + 4 = K e^x \]

### Step 8: Substitute back for \( b \)
Now, substitute back the value of \( b \):
\[ 4x + y + 1 + 4 = K e^x \]
This simplifies to:
\[ 4x + y + 5 = K e^x \]

### Step 9: Solve for \( y \)
Finally, we can solve for \( y \):
\[ y = K e^x - 4x - 5 \]

### Conclusion
Thus, the suitable substitution for solving the differential equation \( \frac{dy}{dx} = 4x + y + 1 \) is:
\[ 4x + y + 1 = b \]