/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 14 Solve the equation. $$ y(x \... [FREE SOLUTION] | 91Ó°ÊÓ

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Solve the equation. $$ y(x \tan x+\ln y) d x+\tan x d y=0. $$

Short Answer

Expert verified
The solution is \(y = A e^{-\frac{x^2}{2}}\), where \(A\) is a constant.

Step by step solution

01

Identify the Type of Differential Equation

The given differential equation is \(y(x \tan x + \ln y) \, dx + \tan x \, dy = 0\). This equation appears to be a non-linear differential equation that is not separable or easily recognizable in a standard form such as exact, homogeneous or linear.
02

Check for Exactness

For a differential equation of the form \(M(x,y) \, dx + N(x,y) \, dy = 0\), the equation is exact if \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\). Here, \(M = y(x \tan x + \ln y)\) and \(N = \tan x\). Compute the partial derivatives:\[\frac{\partial M}{\partial y} = x \tan x + \ln y + 1\]\[\frac{\partial N}{\partial x} = \sec^2 x\]Since \(\frac{\partial M}{\partial y} eq \frac{\partial N}{\partial x}\), the equation is not exact.
03

Use Substitution to Simplify

To simplify, consider substitution techniques. Let \(v = \ln y\), then \(y = e^v\) and \(dy = e^v \, dv\). Substitute these into the equation:\[e^v(x \tan x + v) \, dx + \tan x e^v \, dv = 0\]Divide through by \(e^v\) to get:\[(x \tan x + v) \, dx + \tan x \, dv = 0\]
04

Separate Variables

Now we can attempt to separate variables:\[\tan x \, dv = -(x \tan x + v) \, dx\]Rearrange terms to:\[dv + \frac{v}{\tan x} \, dx = -x \, dx\]Integrate both sides.
05

Integrate Both Sides

Integrate the left side with respect to \(v\) and the right side with respect to \(x\):\[\int \left( dv + \frac{v}{\tan x} \, dx \right) = \int -x \, dx\]This gives:\[v + \int \frac{v}{\tan x} \, dx = -\frac{x^2}{2} + C\]Thus,\[v = -\frac{x^2}{2} + C\]
06

Back-Substitute for y

Recall \(v = \ln y\), substitute back to get:\[\ln y = -\frac{x^2}{2} + C\]Exponentiate both sides to solve for \(y\):\[y = e^{C}e^{-\frac{x^2}{2}}\]Where \(C\) is an integration constant, we can also write \(e^C\) as a new constant \(A\):\[y = A e^{-\frac{x^2}{2}}\].

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Key Concepts

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

Non-linear Differential Equations
Non-linear differential equations are those where the derivatives are raised to a power or multiplied together, or where the function and its derivatives are not in a linear combination. In the given exercise, the equation involves terms like \(x \tan x\) and \(\ln y\), leading to a non-linear equation.

These equations are more complex than linear ones and often require advanced techniques to solve. They cannot generally be expressed in the form \(a(x)y'' + b(x)y' + c(x)y = f(x)\), the standard form for linear differential equations. Instead, they might involve products or transcendental functions such as \(\tan x\).
  • Non-linearity arises from such combinations.
  • Solutions are not straightforward and often need symbolic manipulations and techniques like substitution.
Tackling non-linear equations requires creativity and a good grasp of various mathematical techniques.
Exact Differential Equations
Exact differential equations are a special type of differential equation where the equation can be expressed in the form \(M(x,y) \, dx + N(x,y) \, dy = 0\), with being exact when \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\).

In these equations, there exists a function \(F(x, y)\) such that its total derivative \(dF = M(x, y) \, dx + N(x, y) \, dy\) equals zero. This means it can be integrated directly.
  • Verify exactness by checking the equality of partial derivatives.
  • If exact, there is a potential function that simplifies integration.

As observed in the step-by-step solution, since \(\frac{\partial M}{\partial y} \, eq \frac{\partial N}{\partial x}\), the given differential equation is not exact, necessitating other methods like substitution to solve it.
Variable Substitution
Variable substitution is a technique used to simplify complex differential equations by transforming them into a more familiar or easier form. This involves replacing variables with functions or new variables that simplify the equation for further analysis or integration.

In the provided equation, the substitution \(v = \ln y\) was made, which simplifies the non-linear elements involved with \(y\).
  • Transforms the original differential equation.
  • Often includes derivatives of the substitution to be considered (e.g., \(dy = e^v \, dv\)).

Upon substitution, the equation became easier to separate variables and integrate, demonstrating variable substitution's power to transform and resolve complex differential equations.
Separation of Variables
Separation of variables is a method for solving differential equations by rewriting them so that each variable is contained in a separate function. This allows for straightforward integration.

For the original exercise, after using substitution, the equation allows for separation as follows: \(\tan x \, dv = -(x \tan x + v) \, dx\). This is then arranged such that similar terms are grouped together, leading to separable integrals over \(v\) and \(x\).
  • This technique simplifies integration significantly.
  • Commonly used where each variable pair independently integrates.
By isolating each variable, this method elegantly reduces the problem complexity, making even non-linear equations more tractable.

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