Solving the Given Differential Equation

We are asked to find the solution of the differential equation:

\(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \frac{{{\rm{y}}\phi '\left( {\rm{x}} \right) - {{\rm{y}}^2}}}{{\phi \left( {\rm{x}} \right)}}\)

Let's rearrange the equation to see if it fits a standard form. We can divide both sides by \(\phi \left( {\rm{x}} \right)\):

\(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \frac{{\rm{y}}}{\phi \left( {\rm{x}} \right)}\phi '\left( {\rm{x}} \right) - \frac{{{{\rm{y}}^2}}}{{\phi \left( {\rm{x}} \right)}}\)

This equation involves \(y^2\) and \(y\phi'(x)\), suggesting it might be a Bernoulli equation or solvable with a substitution.

Transforming to a Linear Differential Equation

Let's rearrange the equation to group terms involving \(y\):

\(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} - \frac{{\phi '\left( {\rm{x}} \right)}}{\phi \left( {\rm{x}} \right)}{\rm{y}} = - \frac{1}{{\phi \left( {\rm{x}} \right)}}{{\rm{y}}^2}\)

This is a Bernoulli equation of the form \(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} + P(x)y = Q(x)y^n\), where \(P(x) = - \frac{{\phi '\left( {\rm{x}} \right)}}{\phi \left( {\rm{x}} \right)}\), \(Q(x) = - \frac{1}{{\phi \left( {\rm{x}} \right)}}\), and \(n=2\).

To solve a Bernoulli equation, we typically divide by \(y^n\) and use the substitution \(v = y^{1-n}\).

Divide the equation by \(y^2\):

\(\frac{1}{{{\rm{y}}^2}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}} - \frac{{\phi '\left( {\rm{x}} \right)}}{\phi \left( {\rm{x}} \right)}\frac{1}{{\rm{y}}} = - \frac{1}{{\phi \left( {\rm{x}} \right)}}\)

Let \(v = y^{1-2} = y^{-1} = \frac{1}{{\rm{y}}}\).

Then, differentiate \(v\) with respect to \(x\) using the chain rule:

\(\frac{{{\rm{dv}}}}{{{\rm{dx}}}} = \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{y}}^{ - 1}}} \right) = - 1 \cdot {{\rm{y}}^{ - 2}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = - \frac{1}{{{\rm{y}}^2}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}}\)

So, \(\frac{1}{{{\rm{y}}^2}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = - \frac{{{\rm{dv}}}}{{{\rm{dx}}}}\).

Substitute \(v\) and \(\frac{{{\rm{dv}}}}{{{\rm{dx}}}}\) into the transformed equation:

\(-\frac{{{\rm{dv}}}}{{{\rm{dx}}}} - \frac{{\phi '\left( {\rm{x}} \right)}}{\phi \left( {\rm{x}} \right)}v = - \frac{1}{{\phi \left( {\rm{x}} \right)}}\)

Multiply by -1 to get the standard linear form \(\frac{{{\rm{dv}}}}{{{\rm{dx}}}} + P_{new}(x)v = Q_{new}(x)\):

\(\frac{{{\rm{dv}}}}{{{\rm{dx}}}} + \frac{{\phi '\left( {\rm{x}} \right)}}{\phi \left( {\rm{x}} \right)}v = \frac{1}{{\phi \left( {\rm{x}} \right)}}\)

Here, \(P_{new}(x) = \frac{{\phi '\left( {\rm{x}} \right)}}{\phi \left( {\rm{x}} \right)}\) and \(Q_{new}(x) = \frac{1}{{\phi \left( {\rm{x}} \right)}}\).

Calculating the Integrating Factor

For a linear first-order differential equation \(\frac{{{\rm{dv}}}}{{{\rm{dx}}}} + P(x)v = Q(x)\), the integrating factor (IF) is given by \(e^{\int P(x) dx}\).

In our case, \(P(x) = \frac{{\phi '\left( {\rm{x}} \right)}}{\phi \left( {\rm{x}} \right)}\).

The integral of \(P(x)\) is:

\(\int \frac{{\phi '\left( {\rm{x}} \right)}}{\phi \left( {\rm{x}} \right)} dx\)

This integral is of the form \(\int \frac{{f'(x)}}{{f(x)}} dx\), which is equal to \(\ln|f(x)|\). So,

\(\int \frac{{\phi '\left( {\rm{x}} \right)}}{\phi \left( {\rm{x}} \right)} dx = \ln\left| {\phi \left( {\rm{x}} \right)} \right|\)

The integrating factor is:

IF = \(e^{\ln\left| {\phi \left( {\rm{x}} \right)} \right|} = \left| {\phi \left( {\rm{x}} \right)} \right|\)

Assuming \(\phi(x)\) does not change sign in the interval of interest, we can use IF = \(\phi \left( {\rm{x}} \right)\).

Solving the Linear Equation

Multiply the linear equation \(\frac{{{\rm{dv}}}}{{{\rm{dx}}}} + \frac{{\phi '\left( {\rm{x}} \right)}}{\phi \left( {\rm{x}} \right)}v = \frac{1}{{\phi \left( {\rm{x}} \right)}}\) by the integrating factor \(\phi \left( {\rm{x}} \right)\):

\(\phi \left( {\rm{x}} \right)\left( {\frac{{{\rm{dv}}}}{{{\rm{dx}}}} + \frac{{\phi '\left( {\rm{x}} \right)}}{\phi \left( {\rm{x}} \right)}v} \right) = \phi \left( {\rm{x}} \right)\frac{1}{{\phi \left( {\rm{x}} \right)}}\)

\(\phi \left( {\rm{x}} \right)\frac{{{\rm{dv}}}}{{{\rm{dx}}}} + \phi '\left( {\rm{x}} \right)v = 1\)

The left side is the derivative of the product \(v \cdot \phi \left( {\rm{x}} \right)\) with respect to \(x\):

\(\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {v\phi \left( {\rm{x}} \right)} \right) = 1\)

Now, integrate both sides with respect to \(x\):

\(\int \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {v\phi \left( {\rm{x}} \right)} \right) dx = \int 1 dx\)

\(v\phi \left( {\rm{x}} \right) = x + c\), where \(c\) is the constant of integration.

Substituting Back to Find y

We used the substitution \(v = \frac{1}{{\rm{y}}}\). Substitute this back into the solution:

\(\frac{1}{{\rm{y}}}\phi \left( {\rm{x}} \right) = x + c\)

Now, solve for \(y\):

\(\phi \left( {\rm{x}} \right) = {\rm{y}}\left( {x + c} \right)\)

\({\rm{y}} = \frac{{\phi \left( {\rm{x}} \right)}}{{x + c}}\)

This is the general solution to the given differential equation.

Comparing with Options

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

• Option 1: \({\rm{y}} = \frac{{\rm{x}}}{{\phi \left( {\rm{x}} \right) + {\rm{c}}}}\)
• Option 2: \({\rm{y}} = \frac{{\phi \left( {\rm{x}} \right)}}{{\rm{x}}} + {\rm{c}}\)
• Option 3: \({\rm{y}} = \frac{{\phi \left( {\rm{x}} \right) + {\rm{c}}}}{{\rm{x}}}\)
• Option 4: \({\rm{y}} = \frac{{\phi \left( {\rm{x}} \right)}}{{{\rm{x}} + {\rm{c}}}}\)

Our solution matches Option 4.

Revision Table

Key steps to solve this differential equation:

• Identify the type of differential equation (Bernoulli in this case).
• Use an appropriate substitution (\(v = 1/y\)) to transform it into a linear first-order equation.
• Calculate the integrating factor for the linear equation.
• Multiply the linear equation by the integrating factor to make the left side a total derivative.
• Integrate both sides to find the solution in terms of the substituted variable.
• Substitute back to express the solution in terms of the original variable \(y\).

Additional Information on Differential Equations

Differential equations are equations that relate a function with its derivatives. They are fundamental in modeling processes across various fields like physics, engineering, biology, and economics.

First-Order Differential Equations: These involve only the first derivative of the unknown function. Common types include:

• Separable equations: Can be written in the form \(f(y) dy = g(x) dx\).
• Linear equations: Can be written in the form \(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} + P(x)y = Q(x)\). Solved using an integrating factor.
• Exact equations: Equations of the form \(M(x, y) dx + N(x, y) dy = 0\) where \(\frac{{\partial M}}{{\partial y}} = \frac{{\partial N}}{{\partial x}}\).
• Bernoulli equations: Can be written in the form \(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} + P(x)y = Q(x)y^n\), where \(n \ne 0, 1\). These can be transformed into linear equations using the substitution \(v = y^{1-n}\).

The constant of integration \(c\) represents the family of solutions to the differential equation. A particular solution can be found if an initial condition (a specific value of \(y\) at a given \(x\)) is provided.