Solve the following differential equations. \(y y^{\prime}-2 y^{2} \cot x=\sin x \cos x\)

Short Answer

Expert verified
The general solution is: \( y^{2} = 2 y \ln|\sin x| + \sin x \cos x + c \).

Step by step solution

01

Rewrite the Equation

The given differential equation is \( y y^{\prime}-2 y^{2} \cot x=\sin x \cos x \). To solve this, we first rewrite it in a more workable form. Let's isolate the derivative term: \( y y^{\prime} = 2 y^{2} \cot x + \sin x \cos x \).
02

Simplify the Equation

Divide the entire equation by \( y \): \( y^{\prime} = 2 y \cot x + \frac{\sin x \cos x}{y} \).
03

Express as a Linear Differential Equation

We can rearrange the terms to: \( y^{\prime} - 2 y \cot x = \frac{\sin x \cos x}{y} \). Notice that this can be solved by separation of variables.
04

Separate Variables

Separate the variables: \( y \frac{dy}{dx} = 2 y \cot x + \sin x \cos x \).Divide by \( y \): \( \frac{dy}{dx} = 2 \cot x + \frac{\sin x \cos x }{y} \).Separate to isolate \( y \): \( y \frac{ dy}{dx} = 2 \cot x \cdot y + \sin x \cos x \cdot \frac{1}{y} \).
05

Integrate Both Sides

Integrate on both sides: Integrate the left side with respect to \( y \) and the right side with respect to \( x \): \(\int y dy = \int 2 \cot x \cdot y dx + \int \frac{\sin x \cos x}{y} dx\).
06

Solve the Integrals

Integrate each term separately: \( \frac{1}{2} y^{2} = 2 y \ln|\sin x| + \frac{1}{2 y} \sin 2x + C \).
07

Simplify the Solution

Now simplify the equation and solve for y, the simplified general solution is: \( y^{2} = 2 y \ln|\sin x| + \sin x \cos x + c \). Rearrange terms to solve for \( y \).

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Key Concepts

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

Separation of Variables
One common method for solving differential equations is the separation of variables technique. This method works best for equations that can be rearranged such that one side includes all terms involving the dependent variable and its derivatives, and the other side includes all terms with the independent variable.
In the given problem, the differential equation is initially written as:
\[ yy' - 2y^2 \text{cot}(x) = \text{sin}(x) \text{cos}(x) \]
To use separation of variables, we rearranged terms to isolate the derivative term, resulting in:
\[ y' = 2y \text{cot}(x) + \frac{\text{sin}(x) \text{cos}(x)}{y} \]
By separating the variables in the above equation, we express it in a form where the variables are isolated on each side:
\[ y \frac{dy}{dx} = 2 \text{cot}(x) y + \frac{\text{sin}(x) \text{cos}(x)}{y} \]
Dividing through by y and further rearranging results in separated variables:
\[ \frac{dy}{dx} = 2 \text{cot}(x) + \frac{\text{sin}(x) \text{cos}(x)}{y} \]
This allows us to integrate each side separately to find the solution.
Key steps in separation of variables include:
  • Isolating the variables on opposite sides of the equation.
  • Ensuring the original differential equation is manipulable into separable form.
  • Careful integration of both sides after separation.
Linear Differential Equations
Linear differential equations are a specific type that can generally be written in the form:
\[ y' + P(x)y = Q(x) \]
These equations are called 'linear' because they involve no powers or products of the dependent variable and its derivatives higher than one. In our case, we start with:
\[ yy' - 2y^2 \text{cot}(x) = \text{sin}(x)\text{cos}(x) \]
After isolating the derivative term and rearranging, we get:
\[ y' - 2y \text{cot}(x) = \frac{\text{sin}(x)\text{cos}(x)}{y} \]
Notice this form is similar to the linear standard form:
\[ y' + P(x)y = Q(x) \], where:
\[ P(x) = -2\text{cot}(x) \] and \[ Q(x) = \frac{\text{sin}(x)\text{cos}(x)}{y} \].
Understanding the form and traits of linear differential equations helps to systematically solve them using methods like integrating factors and direct integration.
Key traits of linear differential equations:
  • Involves no products or powers of y and its derivatives.
  • Can often be solved using specific techniques tailored for linear equations like integrating factors.
  • General solutions often involve integrals and constants of integration that need to be solved explicitly.
Integration Techniques
Solving differential equations often involves integration, as it is the process used to find the antiderivative of functions. In our solution:
Integrate each term separately:
\[ \frac{1}{2} y^2 = 2y \text{ln}|\text{sin}(x)| + \frac{1}{2y} \text{sin}(2x) + C \]
Integration techniques used in solving differential equations include:
  • Direct integration when the integrand is straightforward.
  • Using trigonometric identities to simplify complex integrands such as \[ \text{sin}(2x) \].
  • Substitution technique, when variables change to simplify the integral form.
In our specific example, after dividing and arranging the terms to use the separation of variables, solving each side involves integrating:
\[ \frac{1}{2} y^2 = \text{Integral of} \big( 2y \text{ln}|\text{sin}(x)| + \frac{\text{sin}(2x)}{2y} \big) + C \]
Simplifying the integrals by recognizing and using identities such as:
\[ \text{sin}(2x) = 2\text{sin}(x)\text{cos}(x) \] and employing partial fraction decomposition when necessary, are key steps to integrate effectively.

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Most popular questions from this chapter

Solve the following differential equations. \(\left(D^{2}-5 D+6\right) y=0\)

(a) Show that $$ \begin{aligned} D\left(e^{a x} y\right) &=e^{a x}(D+a) y \\ D^{2}\left(e^{a x} y\right) &=e^{a x}(D+a)^{2} y \end{aligned} $$ and so on, that is, for any positive integral \(n_{y}\) $$ D^{n}\left(e^{a x} y\right)=e^{a x}(D+a)^{n} y $$ Thus show that if \(L(D)\) is any polynomial in the operator \(D\), then $$L(D)\left(e^{a x} y\right)=e^{\alpha x} L(D+a) y$$ This is called the exponential shift.

Show that the thickness of the ice on a lake increases with the square root of the time in cold weather, making the following simplifying assumptions. Let the water temperature be a constant \(10^{\circ} \mathrm{C}\), the air temperature a constant \(-10^{\circ} \mathrm{C}_{7}\) and assume that at any given time the ice forms a slab of uniform thickness \(x\). The rate of formation of ice is proportional to the rate at which heat is transferred from the water to the air. Let \(t=0\) when \(x=0\).

Identify each of the differential equations as to type (for example, separable, linear first order, linear second order, etc.), and then solve it. \(\left(x \cos y-e^{-\sin y}\right) d y+d x=0\)

(a) Consider a light beam traveling downward into the ocean. As the beam progresses, it is partially absorbed and its intensity decreases. The rate at whieh the intensity is decreasing with depth at any point is proportional to the intensity at that depth. The proportionality constant \(\mu\) is called the linear absorption coefficient. Show that if the intensity at the surface is \(I_{0}\), the intensity at a distance \(s\) below the surface is \(l=\) \(I_{0} e^{-\mu s} .\) The linear absorption coefficient for water is of the order of \(10^{-2} \mathrm{ft}^{-1}\) (the exact value depending on the wavelength of the light and the impurities in the water). For this value of \(\mu\), find the intensity as a fraction of the surface intensity at a depth of \(1 \mathrm{ft}, 50 \mathrm{ft}, 500 \mathrm{ft}, 1\) mile. When the intensity of a light beam has been reduced to half. its surface intensity \(\left(I=\frac{1}{2} I_{0}\right)\), the distance the light has penetrated into the absorbing substance is called the half-value thickness of the substance. Find the half-value thickness in terms of \(\mu\). Find the half-value thickness for water for the value of \(\mu\) given above. (b) Note that the differential equation and its solution in this problem are mathematically the same as those in Example I, although the physical problem and the terminology are different. In discussing radioactive decay, we call \(\lambda\) the decay constant, and we define the half-life \(T\) of a radioactive substance as the time when \(N=\frac{1}{2} N_{0}\) (compare half-value thickness). Find the relation between \(\lambda\) and \(T\).

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