Question

Find the solution to the initial-value problem $$ 6 y^{\prime}=(2 x+1)\left(y^{2}-2 y-8\right), \quad y(0)=-3 $$ using the method of separation of variables.

Short answer

The solution is \(y = \frac{4-7(2+y)e^{x^2+x}}{1+7e^{x^2+x}}\).

Step-by-step solution

Separate the Variables

The given differential equation is \(6 y' = (2x + 1)(y^2 - 2y - 8)\). We want to separate the variables by moving all terms involving \(y\) to one side and all terms involving \(x\) to the other side. Rewriting the equation yields:\[y' = \frac{(2x + 1)(y^2 - 2y - 8)}{6}\]Thus, we have:\[\frac{dy}{y^2 - 2y - 8} = \frac{(2x + 1)}{6}dx\]

Factor the Denominator

The denominator \(y^2 - 2y - 8\) needs to be factored to simplify the integration process:\[y^2 - 2y - 8 = (y - 4)(y + 2)\]After factoring, the separated equation is:\[\frac{dy}{(y-4)(y+2)} = \frac{(2x + 1)}{6}dx\]

Integrate Both Sides

Now, we need to integrate both sides. Begin with the left side which requires partial fraction decomposition:\[\frac{1}{(y-4)(y+2)} = \frac{A}{y-4} + \frac{B}{y+2}\]Solving for \(A\) and \(B\), we find \(A = \frac{1}{6}\) and \(B = -\frac{1}{6}\). Therefore, the integral can be written as:\[\int \left( \frac{1}{6(y-4)} - \frac{1}{6(y+2)} \right) dy\]Integrate each term:\[\frac{1}{6} \ln|y-4| - \frac{1}{6} \ln|y+2|\]For the right side, integrate the simple polynomial:\[\int \frac{(2x + 1)}{6} dx = \frac{1}{6} \left( x^2 + x \right)\]

Solve for the Constant

Combine the integrated terms and solve for the constant \(C\):\[\frac{1}{6} \ln \frac{|y-4|}{|y+2|} = \frac{1}{6} (x^2 + x) + C\]Multiplying through by 6 and applying the initial condition \(y(0) = -3\):\[\ln \frac{|y-4|}{|y+2|} = x^2 + x + C_1\]Substitute \(x = 0\) and \(y = -3\) into the equation:\[\ln \frac{|-3-4|}{|-3+2|} = C_1 \, \rightarrow \, \ln 7 = C_1\]

Solve for y

With \(C_1 = \ln 7\), insert into the equation:\[\ln \frac{|y-4|}{|y+2|} = x^2 + x + \ln 7\]Exponentiate both sides to remove the logarithm:\[\frac{|y-4|}{|y+2|} = 7e^{x^2 + x}\]Considering the initial condition and knowing \(y = -3\) implies \(y-4 < 0\) and \(y+2 < 0\), we solve for \(y\):\[|y-4| = -(y-4) \quad \text{and} \quad |y+2| = -(y+2)\]Equation becomes:\[\frac{4-y}{2+y} = 7e^{x^2+x}\]Solving for \(y\) gives:\[y = \frac{4 - 7(2+y)e^{x^2+x}}{1 + 7e^{x^2+x}}\]

Result

Thus, the function \(y(x)\) is given as:\[y = \frac{4 - 7(2 + y)e^{x^2 + x}}{1 + 7e^{x^2 + x}}\]

Key concepts

Initial Value Problem

An initial value problem is a differential equation along with a specified value, called the initial condition, which the solution must satisfy. The problem is typically expressed as a differential equation paired with an initial condition, noted as \(y(x_0) = y_0\).
For instance, in the equation \(6 y^{\prime}=(2 x+1)(y^{2}-2 y-8)\), the initial condition is \(y(0) = -3\).
This specific value helps to determine a unique solution amongst the many possible solutions of the differential equation.
In practice, solving an initial value problem involves integrating the differential equation and then applying the initial condition to solve for the constant of integration.
This process is crucial in ensuring that the resulting solution curve passes through the specified point in the coordinate plane that represents the initial condition.

Integration

Integration is a fundamental process in calculus used to determine the antiderivative or 'area' under a curve.
In the context of solving differential equations, integration helps in finding a function whose derivative is a specified function.
This process is the reverse of differentiation.
In our exercise, once the differential equation was separated into two integral forms, each side was independently integrated.

• The left side required partial fraction decomposition to simplify the process for easier integration.
• The right side involved a simple polynomial that follows basic integration rules.

The integration of the polynomial on the right side resulted in \(\frac{1}{6} (x^2 + x)\), demonstrating the straightforward process of integrating polynomials.
The ultimate aim of integration in this scenario is to arrive at functions of x and y that can be solved to express y explicitly in terms of x.

Partial Fraction Decomposition

Partial fraction decomposition is a technique used in calculus to break down complex rational expressions into simpler ones, making them easier to integrate.
It is particularly useful when dealing with fractions where the denominator can be factored into linear or quadratic terms.
In the given exercise, the left side of the differential equation involved the expression: \(\frac{1}{(y-4)(y+2)}\).
Using partial fraction decomposition, this expression was dissected into: \(\frac{A}{y-4} + \frac{B}{y+2}\).

• Solving for A and B provided specific constants that simplified the integration process.
• This breakdown allows straightforward integration of separate terms like \(\frac{1}{y-4}\) and \(\frac{1}{y+2}\).

This method highlights the advantage of reducing complicated expressions into simpler integral terms, thus facilitating a more accessible solution for the associated integrals.

Differential Equations

Differential equations are mathematical equations that involve derivatives, representing rates of change.
They are crucial in modeling various physical phenomena, depicting how a variable changes over time or in relation to other variables.
In our exercise, the differential equation \(6 y^{\prime} = (2x + 1)(y^2 - 2y - 8)\) contained both a derivative \(y^{\prime}\) and polynomial expressions, indicative of how y changes concerning x.
Solving such equations typically involves finding the general form of y(x) that obeys the given rate of change constraints.

• Separation of variables was initially used to isolate terms involving y and x.
• The resulting integrals were solved to provide a combined expression for \(y\) in terms of \(x\).

Differential equations thus serve as fundamental tools in understanding and predicting dynamic systems, making them integral to fields such as physics, engineering, and beyond.