Question

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

Short answer

Based on the given solution, explain why the ODE is nonlinear. The given ODE is nonlinear because the equation involves a product of the dependent variable y and its derivative y', which is represented by the term y y'. In a linear ODE, the dependent variable and its derivatives appear only in linear terms without any multiplication or higher powers. The presence of the term y y' makes the ODE nonlinear.

Step-by-step solution

Separation of Variables

We first manipulate the given equation to make it easier to find the function y(x). Divide both sides of the equation by x to isolate the derivative term: $$ y y'+ \frac{x}{y}+ y=0 $$ Rearrange the terms to get: $$ y y'=-\frac{x}{y}- y $$ Observe that the equation is now separated in terms of y and x: $$ \frac{y'}{-\frac{x}{y^2}- \frac{y}{x}}=1 $$

Integrate both sides

Now we want to integrate both sides of the equation with respect to x. Note that in order to do this integration we need to recognize that the derivative we are integrating against is \(y'\), the notation for the left side integral should be \(\int \mathrm{d}y\): $$ \int \frac{y^2}{x+y^2} \mathrm{d}y = \int \mathrm{d}x $$ For clarity, let's denote the antiderivative of the left side as F(y) and the antiderivative of the right side as G(x): $$ F(y) = G(x) + C $$ Where C is the constant of integration.

Integrate the left side

To integrate the left side, perform the substitution \(u= x + y^2\) which gives \(\mathrm{d}u = 2y\mathrm{d}y\). Then, we have: $$ \int \frac{y^2\mathrm{d}y}{x+y^2} = \frac{1}{2}\int \frac{u- x}{u} \mathrm{d}u = \frac{1}{2}\int (1- \frac{x}{u}) \mathrm{d}u $$ Now, perform the integration: $$ \frac{1}{2}(\int 1 \mathrm{d}u - \int \frac{x}{u} \mathrm{d}u)= \frac{1}{2}(u- x\ln{u}) $$ Substitute back y to find F(y): $$ F(y) = \frac{1}{2}(x+y^2- x\ln{(x+ y^2)}) $$

Integrate the right side and find the solution

Now let's find G(x) by integrating the right side: $$ G(x) = \int \mathrm{d}x = x $$ Now, equate F(y) and G(x) and add the constant of integration: $$ \frac{1}{2}(x + y^2 - x\ln{(x+y^2)}) = x+C $$ Now we apply the initial condition y(1) = 2 to find C: $$ \frac{1}{2}(1 + 4 - 1\ln{(1+4)}) = 1+C \\ C =-\frac{1}{2} $$ Now, we have the implicit solution to the initial value problem: $$ \frac{1}{2}(x+y^2- x\ln{(x+y^2)}) = x-\frac{1}{2} $$ This is the final solution to the initial value problem, and it can't be simplified further. Although the solution is not in the explicit form, y(x), it still satisfies the given equation and the initial condition.

Key concepts

Separation of Variables

The method of separation of variables is a widely used technique for solving differential equations, particularly when dealing with initial value problems. It involves reorganizing an equation so that each variable appears on a different side of the equation, allowing for independent integration with respect to each variable. For the given exercise, the process starts by dividing the original equation by x to isolate the derivative term, and then rearranging it to explicitly separate variables associated with y and x.

The resulting equation is ready for the integration step. This method is applicable when the differential equation can be manipulated into a multiplicative separation of variables, which is essentially transforming the problem into a form where the solution to the differential equation can be expressed as the product of two functions, each depending on a single variable. In some cases, additional substitutions might be necessary to achieve complete separation.

Integrating Differential Equations

Once we have separated the variables, as seen in the separation of variables step, the next challenge is integrating the equation to find the solution. Integration of differential equations is a crucial step where we find the antiderivatives of both sides of the separated equation. This often involves recognizing the correct substitution to simplify the integrals, as not all integrals can be directly computed. For instance, in the given problem,

\[\[\begin{align*}
F(y) &= \frac{1}{2}(x+y^2- x\text{ln}(x+ y^2))
\text{and}
G(x) &= x
\text{are determined by integrating the separated terms.}\end{align*}\]\]

The process includes identifying integration techniques such as substitution, integration by parts, partial fractions, or sometimes numerical methods if the integral cannot be expressed in terms of elementary functions. After integrating both sides, we include the constant of integration, which will later be determined using the given initial condition.

Boundary Value Problems

Boundary value problems (BVPs) involve differential equations where the solution is sought under specific conditions at the boundaries of the domain. While our exercise deals with an initial value problem, understanding BVPs is also important in the mathematical analysis of physical systems. BVPs differ from initial value problems in that the conditions are specified at more than one point, often at the endpoints of an interval.

For example, in the context of physical problems, BVPs can describe the steady-state distribution of temperatures along a rod where the temperatures at both ends are held fixed. Solving BVPs typically involve methods like the shooting method, finite difference method, or transforming the problem into an equivalent eigenvalue problem. Understanding the concepts behind BVPs helps in grasping the broader scope of solving differential equations under various constraints.