Question

Solve the initial value problem. $$ y^{\prime}+x\left(y^{2}+y\right)=0, \quad y(2)=1 $$

Short answer

Question: Determine the implicit solution of the initial value problem for the given non-linear first-order differential equation and initial condition. Differential equation: $$\frac{dy}{dx} + x(y^2 + y) = 0$$ Initial condition: $$y(2) = 1$$ Answer: The implicit solution for the given initial value problem is: $$y(x) = (y(x)+1)e^{-\frac{1}{2}x^2 + 1}$$

Step-by-step solution

Rewrite the Equation in Separable Form

To rewrite the given differential equation in separable form, we can follow these steps: 1. Divide by (y^2 + y) to separate the variables: $$ \frac{dy}{dx} = -x(y^2 + y) $$ 2. Move the terms with y to the left side, and those with x to the right side: $$ \frac{dy}{(y^2 + y)} = -x\, dx $$ Now, we have a separable differential equation.

Integrating Both Sides

Now, we integrate both sides of the equation: $$ \int \frac{dy}{(y^2 + y)} = \int (-x\, dx) $$ To integrate the left side, let's use partial fraction decomposition. Rewrite the integrand as follows: $$ \frac{1}{y^2 + y} = \frac{A}{y} + \frac{B}{y+1} $$ Solving for A and B, we have A = 1 and B = -1. So, the left side becomes: $$ \int \frac{dy}{(y^2 + y)} = \int (\frac{1}{y} - \frac{1}{y+1})\, dy $$ Now, we are ready to integrate: $$ \int (\frac{1}{y} - \frac{1}{y+1})\, dy = -\int x\, dx $$

Evaluate the Integrals

Now, we evaluate the integrals: $$ \left[\ln|y| - \ln|y+1|\right] = -\frac{1}{2}x^2 + C $$ Apply the property of logarithms to simplify the equation: $$ \ln\left|\frac{y}{y+1}\right| = -\frac{1}{2}x^2 + C $$ Next, we exponentiate both sides to get rid of the logarithm: $$ \frac{y}{y+1} = e^{-\frac{1}{2}x^2 + C} $$

Solving for y and Apply Initial Condition

We can rewrite the last equation as: $$ y = (y+1)e^{-\frac{1}{2}x^2 + C} $$ Now we apply the initial condition y(2) = 1: $$ 1 = (1+1)e^{-\frac{1}{2}(2)^2 + C} $$ Solving for C, we find that C = 1. Thus, the equation becomes: $$ y = (y+1)e^{-\frac{1}{2}x^2 + 1} $$

Solving for y (Implicit Solution)

At this point, it is difficult to isolate y on one side of the equation. However, we can express y implicitly as follows: $$ y(x) = (y(x)+1)e^{-\frac{1}{2}x^2 + 1} $$ This implicit form represents the solution to the initial value problem.

Key concepts

Separable Differential Equations

Separable differential equations form a subclass of ordinary differential equations (ODEs), which are especially approachable because we can 'separate' the variables on either side of the equation. To solve such an equation, we rewrite it in a form where all terms containing one variable (typically, the dependent variable, like y) are on one side, and all terms involving the other variable (the independent variable, like x) are on the other side.

Once we do this, we can integrate both sides with respect to their own variables. This approach transforms the problem from solving a differential equation to solving two separate integral equations. In essence, the integration process helps us find a function (or set of functions) aligned with the initial equation. Importantly, separable equations can often model naturally occurring phenomena allowing for their prediction and understanding. An initial value, like in our exercise where y(2)=1, gives us a specific point to 'start' from, enabling us to find a unique solution that corresponds to this point.

Partial Fraction Decomposition

Partial fraction decomposition is a useful technique in integration, particularly when we deal with rational expressions. A rational expression is a fraction where the numerator and the denominator are both polynomials. When the degree of the numerator is less than the denominator, we can sometimes express the rational expression as a sum of simpler fractions, which we call 'partial fractions.'

Here's what makes partial fraction decomposition handy: it transforms complex fractions into a series of simpler fractions that are easier to integrate. In our case, the complex fraction \(\frac{1}{y^2 + y}\) was decomposed into \(\frac{1}{y} - \frac{1}{y+1}\), where the constants A and B are chosen so that when the right side is combined, it equals the original fraction.

Once we have the partial fractions, we can tackle each simpler integral separately. This step-by-step breakdown not only makes the problem more manageable but also allows us to employ fundamental integration techniques on each piece.

Integration Techniques

Integration, the process of finding integrals, is a foundational tool in calculus. It's essentially the reverse of differentiation, and it has a wide array of applications, from computing areas under curves to solving differential equations. There are several key techniques for integration, each suited to different types of functions or situations.

For instance, we can use substitution when a function is the derivative of another function nested inside a third function—this is like unwrapping layers. We might employ partial fractions, as we did in the exercise, when dealing with rational functions to break them down into simpler parts. And for trigonometric integrals, we often rely on specific identities that simplify the integral.

In our example, the step-by-step solution uses partial fraction decomposition followed by the integration of simple logarithmic functions. This highlights how combining multiple integration techniques can effectively solve more complex problems. Always remember that a strong grasp of these techniques is vital for tackling integrals that arise both in theoretical calculations and real-world applications.