Question

The solution of $e^{\frac{dy}{dx}} = x + 1, \, y(0) = 3$ is:

• A. $y - 2 = x\log x$
• B. $y - x - 3 = x\log x$
• C. $y - x - 3 = (x + 1)\log(x + 1)$
• D. $y + x - 3 = (x + 1)\log(x + 1)$

Answer 1

The Correct Option is D

1. Rewrite the differential equation:

Given: \( e^{\frac{dy}{dx}} = x + 1 \)

Take natural logarithm of both sides:

\[ \frac{dy}{dx} = \ln(x + 1) \]

2. Integrate both sides:

\[ y = \int \ln(x + 1) \, dx \]

Use integration by parts (\( u = \ln(x+1) \), \( dv = dx \)):

\[ y = (x + 1)\ln(x + 1) - \int \frac{x + 1}{x + 1} \, dx = (x + 1)\ln(x + 1) - x + C \]

3. Apply the initial condition \( y(0) = 3 \):

\[ 3 = (0 + 1)\ln(0 + 1) - 0 + C \implies 3 = 0 + C \implies C = 3 \]

Thus:

\[ y = (x + 1)\ln(x + 1) - x + 3 \]

Rearrange:

\[ y + x - 3 = (x + 1)\ln(x + 1) \]

Correct Answer: (D) \( y + x - 3 = (x + 1) \log (x + 1) \)

Answer 2

The given differential equation is: \[ e^{\frac{dy}{dx}} = x + 1. \] Take the natural logarithm on both sides: \[ \frac{dy}{dx} = \ln(x + 1). \] Integrate both sides with respect to $x$: \[ y = \int \ln(x + 1) \, dx + C. \] Using integration by parts: \[ \int \ln(x + 1) \, dx = (x + 1)\ln(x + 1) - \int \frac{x + 1}{x + 1} \, dx = (x + 1)\ln(x + 1) - (x + 1). \] Thus: \[ y = (x + 1)\ln(x + 1) - (x + 1) + C. \] Apply the initial condition $y(0) = 3$: \[ 3 = (0 + 1)\ln(0 + 1) - (0 + 1) + C \implies C = 3. \] The solution becomes: \[ y = (x + 1)\ln(x + 1) - (x + 1) + 3 \implies y + x - 3 = (x + 1)\ln(x + 1). \]

Answer 3

$$ e^{\frac{dy}{dx}} = x + 1 $$

Take the natural logarithm of both sides:

$$ \frac{dy}{dx} = \ln(x + 1) $$

Now, separate the variables:

$$ dy = \ln(x + 1) \, dx $$

Integrate both sides of the equation:

$$ \int dy = \int \ln(x + 1) \, dx $$

The left side is easy:

$$ y = \int \ln(x + 1) \, dx $$

To integrate $ \ln(x + 1) $, use integration by parts:

Let $ u = \ln(x + 1) $ and $ dv = dx $.
Then $ du = \frac{1}{x + 1} \, dx $ and $ v = x $.

Integration by parts formula: $ \int u \, dv = uv - \int v \, du $

$$ \int \ln(x + 1) \, dx = x \ln(x + 1) - \int \frac{x}{x + 1} \, dx $$

To solve the remaining integral, perform polynomial long division or notice that $ x = (x+1) - 1 $:

$$ \int \frac{x}{x + 1} \, dx = \int \frac{(x + 1) - 1}{x + 1} \, dx = \int \left[1 - \frac{1}{x + 1}\right] \, dx = x - \ln|x + 1| + C $$

Substitute this back into the equation for $ y $:

$$ y = x \ln(x + 1) - x + \ln|x + 1| + C $$

Now, apply the initial condition $ y(0) = 3 $:

$$ 3 = 0 \cdot \ln(1) - 0 + \ln(1) + C $$ $$ 3 = C $$

Therefore, the solution is:

$$ y = x \ln(x + 1) - x + \ln(x + 1) + 3 $$

Rearranging gives:

$$ y + x - 3 = (x + 1)\ln(x + 1) $$

This corresponds to option (D).