Question

Solve the following differential equations by using integrating factors. $$ \sin (x) y^{\prime}=y+2 x $$

Short answer

The solution is \( y = \frac{x^2 + C}{\csc(x) - \cot(x)} \).

Step-by-step solution

Rewrite the Equation

First, we need to rewrite the given differential equation in standard linear form. The standard form is:\[ y' + P(x)y = Q(x) \]Our equation is \( \sin(x) y' = y + 2x \). To match this with the standard form, divide all terms by \( \sin(x) \):\[ y' - \frac{1}{\sin(x)} y = \frac{2x}{\sin(x)} \]

Identify Integrating Factor

The integrating factor \( \mu(x) \) is found with the formula:\[ \mu(x) = e^{\int P(x) \, dx} \]Here, \( P(x) = -\frac{1}{\sin(x)} \). First, compute:\[ \int -\frac{1}{\sin(x)} \, dx \]This simplifies to:\[ \int -\csc(x) \, dx = \ln(\csc(x) - \cot(x)) \]Thus, the integrating factor \( \mu(x) \) is:\[ \mu(x) = \csc(x) - \cot(x) \]

Multiply Through by the Integrating Factor

Multiply the entire differential equation by the integrating factor \( \mu(x) = \csc(x) - \cot(x) \) to make the left-hand side an exact derivative:\[ (\csc(x) - \cot(x))y' - (\frac{\csc(x) - \cot(x)}{\sin(x)})y = (\csc(x) - \cot(x))\frac{2x}{\sin(x)} \] This equation simplifies as the left-hand side is the derivative of \( y(\csc(x) - \cot(x)) \).

Integrate Both Sides

Integrate both sides of the equation with respect to \( x \):\[ \int d[y(\csc(x) - \cot(x))] = \int (2x) \, dx \]The left side becomes \( y(\csc(x) - \cot(x)) \), and the right integrates to \( x^2 + C \). Thus:\[ y(\csc(x) - \cot(x)) = x^2 + C \]

Solve for \( y \)

Rearrange the equation from Step 4 to solve for \( y \):\[ y = \frac{x^2 + C}{\csc(x) - \cot(x)} \] This equation represents the solution to the differential equation.

Key concepts

Integrating Factor

The integrating factor is a powerful tool for solving linear differential equations. It simplifies the process by transforming a non-exact equation into an exact one. To find the integrating factor, identify functions that allow the differential equation to be written in the form:

• Multiply the entire differential equation by a suitable function, called the integrating factor.
• This function has the special property that it converts the left-hand side of the equation into the derivative of a product of functions.

For linear first-order differential equations like the ones we deal with here, the integrating factor is given by:\[ \mu(x) = e^{\int P(x) \, dx} \]where \( P(x) \) is the coefficient of \( y \) in the standard form of the equation.
Calculating this integral can include various techniques, such as substitution or partial fractions, depending on its complexity. Once you multiply the whole equation by this integrating factor, you are able to set the stage for solving it more easily.

Linear Differential Equations

Linear differential equations are an important class of equations in calculus with specific characteristics. They take the form:

• Involves derivatives of the unknown function and no higher powers of the function or its derivatives.
• The coefficients of the function and its derivatives are allowed to be functions of the independent variable only.

An example of the standard linear form of a first-order differential equation is:\[ y' + P(x)y = Q(x) \]
In the given problem, we start with \( \sin(x) y' = y + 2x \) which is rearranged by dividing through by the trigonometric term \( \sin(x) \), making it easier to manipulate into the required form. Understanding these forms and transformations is key to finding an integrating factor and a subsequent solution.

Exact Solutions

When dealing with differential equations, exact solutions provide a precise answer to the function accounting for all given conditions. Achieving an exact solution often involves the converting process performed by the integrating factor.
Once you have an expression that can be integrated, finding the particular solution of the differential equation becomes straightforward:

• The operation involves integrating the left side of the equation directly.
• This yields the expression \( y(\csc(x) - \cot(x)) = x^2 + C \), where \( C \) is an integration constant.

Finally, solving for \( y \) gets you to the desired exact solution: \( y = \frac{x^2 + C}{\csc(x) - \cot(x)} \). This demonstrates how the method systematically transforms complex expressions into comprehensible forms.

Calculus Techniques

The procedures of solving differential equations often involve various calculus techniques to simplify and solve expressions. Here, we employed:

• Integration: Once you have the derivative as a single entity, integrate both sides to find the unknown function.
• Substitution: This is useful when dealing with non-standard integrals and can simplify the process of calculating integrals.
• Rearranging and simplifying: Basic algebraic techniques help in bringing the equation to a form where calculus techniques can be efficiently applied.

These techniques, combined with understanding of differential properties and transformations, allow solving even complicated equations. Recognizing when and how to apply them is essential for clear and easy solutions.