Chapter 8: Problem 24

Identify each of the differential equations as type (for example, separable, linear first order, linear second order, etc.), and then solve it. $$x\left(y y^{\prime \prime}+y^{\prime 2}\right)=y y^{\prime} \quad \text { Hint: Let } u=y y^{\prime}$$.

Short Answer

Expert verified

The differential equation is a second-order non-linear differential equation and the solution after integration is \[ y y' = C' x^{y}\].

Step by step solution

- Identifying the Differential Equation Type

The given differential equation is \[x(y y'' + y'^2) = y y'\]. This equation involves second-order and first-order derivatives, suggesting it is a second-order non-linear differential equation.

- Substitution with Suggested Hint

The hint suggests letting: $ u = y y' $. Differentiate this substitution with respect to x: \[u' = (y y')' = y' y' + y y'' = y'^{2} + y y''\]. Thus, the differential equation simplifies to: \[x u' = y u\].

- Separate Variables

Rearrange the equation to separate the variables:\[u' = \frac{u y}{x}\]. By letting $u = y y'$, the equation becomes: \[\frac{du}{u} = \frac{y dx}{x}\].

- Integrate Both Sides

Integrate both sides of the equation:\[\int \frac{1}{u} du = \int \frac{y}{x} dx\]. This gives: \[ \ln |u| = y \ln |x| + C\], where C is the integration constant.

- Simplify the Solution

Exponentiate both sides to solve for u:\[ |u| = e^{y \ln |x| + C} = e^C e^{y \ln |x|} = C' x^y \]. Hence, \[ u = C' x^y \], where $ C' = e^C $.

- Back Substitution

Substitute back $ u = y y' $: \[ y y' = C' x^y \].

- Solve the Differential Equation

To solve $ y y' = C' x^y $, separate variables: \[ y \frac{dy}{y} = C' x^{y - 1} dx \]. Integrate both sides to find the general solution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Second-Order Differential Equations

Second-order differential equations involve the second derivative of a function. They often appear in various fields such as physics, engineering, and economics. The general form is $ y'' + p(x) y' + q(x) y = g(x) $. The given equation $ x(y y'' + y'^2) = y y' $ involves second and first derivatives, making it clearly a second-order equation. In our problem, using the hint \ u = y y' \ simplifies our differential equation into something more manageable. Understanding these concepts is crucial for solving complex problems step by step.

Non-Linear Differential Equations

Non-linear differential equations have terms that are not purely linear, like $y y'' + y'^2$. They often describe real-world systems more accurately, such as weather models or fluid dynamics. The complexity arises because the solution techniques differ from linear ones. In the given problem, the equation $ x(y y'' + y'^2) = y y' $ is non-linear due to products of the function and its derivatives. The trick is to make substitutions that simplify the equation into a solvable form, as shown with \ u = y y'\. Solving non-linear differential equations often requires clever manipulations or approximations.

Separation of Variables

Separation of variables is a method used to solve differential equations by expressing the equation such that each variable appears on one side. For example, our aim with $ x(y y'' + y'^2) = y y' $ was to rearrange it to \ $ u' = \frac{u y}{x} $ after substitution. Once we separate variables, we integrate each side independently. This method is especially useful for tackling equations that initially look too complicated. In our problem, after the substitution, we got to an integrable form $ \int \frac{1}{u} du = \int \frac{y}{x} dx$. Finally, remember to substitute back any changes to get the final solution.