Solving the Differential Equation: A Step-by-Step Guide

We are asked to find the solution of the differential equation \((1 + 3x)dy - (1 - 3y)dx = 0\) with the initial condition \(y(1) = 0\). This is a first order differential equation, and we can solve differential equation using the method of separation of variables.

Separating the Variables

First, let's rearrange the given equation to separate the terms involving \(y\) and \(dy\) from the terms involving \(x\) and \(dx\):

The given equation is:

\((1 + 3x)dy - (1 - 3y)dx = 0\)

Move the dx term to the right side:

\((1 + 3x)dy = (1 - 3y)dx\)

Now, divide both sides by \((1 + 3x)\) and \((1 - 3y)\) to separate the variables:

\(\frac{dy}{1 - 3y} = \frac{dx}{1 + 3x}\)

Note that this separation is valid as long as \(1 - 3y \ne 0\) and \(1 + 3x \ne 0\). The initial condition \(y(1) = 0\) means \(1 - 3(0) = 1 \ne 0\) at \(x=1, y=0\), so we proceed.

Integrating Both Sides to Find the General Solution

To find the general solution, we integrate both sides of the separated equation:

\(\int \frac{dy}{1 - 3y} = \int \frac{dx}{1 + 3x}\)

Let's evaluate the integrals separately.

For the left side, \(\int \frac{dy}{1 - 3y}\), we can use a substitution. Let \(u = 1 - 3y\). Then \(du = -3dy\), which means \(dy = -\frac{1}{3}du\). The integral becomes:

\(\int \frac{-\frac{1}{3}du}{u} = -\frac{1}{3} \int \frac{1}{u} du = -\frac{1}{3} \ln|u| + C_1 = -\frac{1}{3} \ln|1 - 3y| + C_1\)

For the right side, \(\int \frac{dx}{1 + 3x}\), we can use a similar substitution. Let \(v = 1 + 3x\). Then \(dv = 3dx\), which means \(dx = \frac{1}{3}dv\). The integral becomes:

\(\int \frac{\frac{1}{3}dv}{v} = \frac{1}{3} \int \frac{1}{v} dv = \frac{1}{3} \ln|v| + C_2 = \frac{1}{3} \ln|1 + 3x| + C_2\)

Equating the results from both sides:

\(-\frac{1}{3} \ln|1 - 3y| + C_1 = \frac{1}{3} \ln|1 + 3x| + C_2\)

Combine the constants \(C = C_2 - C_1\):

\(-\frac{1}{3} \ln|1 - 3y| = \frac{1}{3} \ln|1 + 3x| + C\)

Multiply by 3:

\(-\ln|1 - 3y| = \ln|1 + 3x| + 3C\)

Let \(C' = 3C\). Rearrange the terms:

\(\ln|1 + 3x| + \ln|1 - 3y| = -C'\)

Using the logarithm property \(\ln a + \ln b = \ln (ab)\):

\(\ln|(1 + 3x)(1 - 3y)| = -C'\)

Exponentiate both sides:

\(|(1 + 3x)(1 - 3y)| = e^{-C'} \)

Since \(e^{-C'}\) is a positive constant, we can remove the absolute value by introducing a constant \(K = \pm e^{-C'}\). If we allow \(K=0\), the trivial solution \(y=1/3\) or \(x=-1/3\) would result, which might or might not satisfy the original equation and boundary condition. For now, let's assume \(K \ne 0\).

\((1 + 3x)(1 - 3y) = K\)

This is the general solution of the first order differential equation.

Using the Initial Condition to Find the Particular Solution

We are given the initial condition \(y(1) = 0\). This means that when \(x = 1\), \(y = 0\). We substitute these values into the general solution to find the value of the constant \(K\).

\((1 + 3(1))(1 - 3(0)) = K\)

\((1 + 3)(1 - 0) = K\)

\((4)(1) = K\)

\(4 = K\)

Now substitute \(K = 4\) back into the general solution to get the particular solution:

\((1 + 3x)(1 - 3y) = 4\)

Expanding and Rearranging the Particular Solution

Let's expand the left side of the equation:

\(1(1 - 3y) + 3x(1 - 3y) = 4\)

\(1 - 3y + 3x - 9xy = 4\)

Rearrange the terms to match the format of the options provided. Let's move the constant term to the right side:

\(3x - 3y - 9xy = 4 - 1\)

\(3x - 3y - 9xy = 3\)

Divide the entire equation by 3:

\(\frac{3x}{3} - \frac{3y}{3} - \frac{9xy}{3} = \frac{3}{3}\)

\(x - y - 3xy = 1\)

This is the particular solution that satisfies the given initial condition \(y(1)=0\).

Comparing with Options

Let's compare our derived particular solution with the given options:

Our derived solution \(x - y - 3xy = 1\) matches Option 3.

To solve differential equation problems like this, identifying the type of first order differential equation and applying the correct method (like variable separable) is key. The initial condition is then used to find the specific particular solution.