Differential Equation Solution Explained

This problem asks us to find the particular solution of a given differential equation that satisfies a specific initial condition. A differential equation is an equation that relates one or more functions and their derivatives. The initial condition helps us find a unique solution among many possible solutions.

Understanding the Given Differential Equation

We are given the differential equation:

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

And the initial condition is \(y(1) = 1\), which means when \(x=1\), the value of \(y\) is \(1\).

Separating Variables for Integration

To solve this first-order differential equation, we can use the method of separation of variables. This involves rearranging the equation so that all terms involving \(y\) and \(dy\) are on one side, and all terms involving \(x\) and \(dx\) are on the other side.

Let's rearrange the given equation:

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

Divide both sides by \(y\) and multiply both sides by \(dx\):

\[\frac{1+y}{y} dy = dx\]

We can simplify the left-hand side:

\[\left(\frac{1}{y} + \frac{y}{y}\right) dy = dx\]

\[\left(\frac{1}{y} + 1\right) dy = dx\]

Integrating Both Sides of the Equation

Now that the variables are separated, we can integrate both sides of the equation. This step will introduce a constant of integration, which we will determine using the initial condition.

Integrate the left side with respect to \(y\) and the right side with respect to \(x\):

\[\int \left(\frac{1}{y} + 1\right) dy = \int dx\]

Performing the integration:

• The integral of \(\frac{1}{y}\) with respect to \(y\) is \(\ln|y|\).
• The integral of \(1\) with respect to \(y\) is \(y\).
• The integral of \(1\) with respect to \(x\) is \(x\).

So, the general solution is:

\[\ln|y| + y = x + C\]

where \(C\) is the constant of integration.

Applying the Initial Condition y(1) = 1

The initial condition \(y(1) = 1\) means that when \(x=1\), \(y=1\). We will substitute these values into our general solution to find the value of \(C\).

Substitute \(x=1\) and \(y=1\) into the equation \(\ln|y| + y = x + C\):

\[\ln|1| + 1 = 1 + C\]

We know that \(\ln(1) = 0\).

\[0 + 1 = 1 + C\]

\[1 = 1 + C\]

Subtracting 1 from both sides gives us:

\[C = 0\]

Final Solution of the Differential Equation

Now that we have found the value of \(C\), we can substitute it back into our general solution to get the particular solution that satisfies the given initial condition.

Substitute \(C=0\) into \(\ln|y| + y = x + C\):

\[\ln|y| + y = x + 0\]

\[\ln|y| + y = x\]

Since \(y(1)=1\), \(y\) is positive, so we can remove the absolute value sign:

\[\ln(y) + y = x\]

To match the form of the options, let's transform this equation. We can exponentiate both sides using the base \(e\):

\[e^{\ln(y) + y} = e^x\]

Using the exponent rule \(e^{A+B} = e^A \cdot e^B\):

\[e^{\ln(y)} \cdot e^y = e^x\]

Since \(e^{\ln(y)} = y\):

\[y \cdot e^y = e^x\]

This is the particular solution to the differential equation that satisfies the given initial condition.

Comparing this result with the provided options:

• Option 1: \((1 + y)e^y = 2e^x\)
• Option 2: \(ye^y = e^x\)
• Option 3: \(y^2e^y = e^x\)
• Option 4: \(2ye^y = e^x + e\)

Our derived solution \(ye^y = e^x\) matches Option 2.