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Solve the differential equation $$ \sin x \frac{d y}{d x}+2 y \cos x=1 $$ subject to the boundary condition \(y(\pi / 2)=1\).

Short Answer

Expert verified
The solution to the given differential equation subject to the boundary condition \(y(\pi / 2)=1\) is y = \(\frac{-\cos x + 1}{\sin^2 x}\)

Step by step solution

01

Determination of the integrating factor

In order to solve the given differential equations, an integrating factor which is contorted by \( e^{\int b(x)/a(x) dx} \) needs to be determined. Simplifying, the integrating factor becomes, \( e^{\int 2\cot x dx} \) = \( e^{2\ln |\sin x|} \) = \( \sin^2 x \)
02

Multiply through by the integrating factor and rearrange

\(\sin x \frac{d y}{d x}+2 y \cos x = 1\) is multiplied by the integrating factor \( \sin^2 x \), resulting in \(\sin^3 x y' + 2\sin^2 x y\cos x = \sin^2 x\). After multiplying through and dividing by \(\sin x\), you get \(\sin^2 x y' + 2\sin x y\cos x = \sin x\) . After rearranging this turns into \(\frac{d}{dx}(\sin^2 x y) = \sin x \).
03

Integrate

Integrating both sides of the equation you get, \(\int d(\sin^2 x y) = \int \sin x dx \). Since, \(\int \sin x dx\) = \(-\cos x + \)constant, it is easy to solve now.
04

Solve for y

Now that the integration process has been completed, solve for Y, \(\sin^2 x y = -\cos x + \)constant, hence y = \(\frac{-\cos x + \)constant}{\(\sin^2 x\)}.
05

Apply boundary condition

The constant value is determined by applying the provided boundary condition \(y(\pi / 2)=1\). So, when x = \(\pi / 2\), y = 1, hence 1 = \(\frac{-\cos(\pi/2) + \)constant}{\(\sin^2(\pi/2)\)} and constant = 1.
06

Final solution

The final solution y is then given by inserting the constant value into the equation y = \(\frac{-\cos x + \)constant}{\(\sin^2 x\)} to obtain y = \(\frac{-\cos x + 1}{\sin^2 x}\).

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Key Concepts

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

Integrating Factor Method
The integrating factor method is a crucial tool for solving differential equations, especially of the linear variety. It helps in transforming non-exact equations into exact ones, facilitating a straightforward integration. The method involves finding a function, usually denoted as the integrating factor, which, when multiplied by the original equation, allows the left-hand side of the equation to be written as the derivative of a product of functions.

In the context of the given exercise, the differential equation \(\sin x \frac{d y}{d x}+2 y \cos x=1\) is first multiplied by an integrating factor determined by the expression \(e^{\int 2\cot x dx}\) which simplifies to \(\sin^2 x\). This factor is precisely what enables us to rewrite the equation as the total derivative of a single function, \(\sin^2 x y\), making it ready for integration. This step is a bridge between setting up the problem and finding a solution through integration.
Boundary Conditions
Boundary conditions are crucial in the realm of differential equations as they allow the determination of the specific solution that corresponds to the physical, geometric, or other constraints of a problem. When tackling a differential equation, there is generally a family of solutions possible, and applying the boundary conditions helps to identify the unique result that fits the case in question.

For the exercise provided, we're given the boundary condition \(y(\pi / 2)=1\). After integrating and finding the general solution, this condition is used to find the particular value of the constant that arises from the integration. By substituting the values of \(x\) and \(y\) from the condition into the general solution, we pinpoint the exact equation representing the scenario in question. The practical role of boundary conditions is to anchor the abstract notion of a solution into a concrete context that makes physical or conceptual sense.
Integration in Calculus
Integration is one of the two fundamental operations in calculus, along with differentiation. While differentiation is concerned with rates of change, integration is focused on accumulation or the total value of a quantity. Solving differential equations often involves integrating, which requires finding the antiderivative or the integral of functions.

In our exercise, integration comes into play in Step 3, where both sides of the equation \(\frac{d}{dx}(\sin^2 x y) = \sin x\) are integrated. The integration of the right side yields \( -\cos x + \) constant, using well-known antiderivatives. Integration serves not just as a mathematical process but also as a conceptual bridge from the rate at which something happens to the quantity or total effect over time, a key understanding in the physical sciences and engineering. This step transforms the problem from its differential form—where we are concerned with the rate of change of \(y\)—into a function that represents the accumulated effect, thus giving us a solution for \(y\) itself.

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Most popular questions from this chapter

Find the family of solutions of $$ \frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2}+\frac{d y}{d x}=0 $$ that satisfy \(y(0)=0\).

The two functions \(x(t)\) and \(y(t)\) satisfy the simultaneous equations $$ \begin{aligned} &\frac{d x}{d t}-2 y=-\sin t \\ &\frac{d y}{d t}+2 x=5 \cos t \end{aligned} $$ Find explicit expressions for \(x(t)\) and \(y(t)\), given that \(x(0)=3\) and \(y(0)=2\). Sketch the solution trajectory in the \(x y\)-plane for \(0 \leq t<2 \pi\), showing that the trajectory crosses itself at \((0,1 / 2)\) and passes through the points \((0,-3)\) and \((0,-1)\) in the negative \(x\)-direction.

If \(u=1+\tan y\), calculate \(d(\ln u) / d y\); hence find the general solution of $$ \frac{d y}{d x}=\tan x \cos y(\cos y+\sin y) $$

For a lightly damped \(\left(\gamma<\omega_{0}\right)\) harmonic oscillator driven at its undamped resonance frequency \(\omega_{0}\), the displacement \(x(t)\) at time \(t\) satisfies the equation $$ \frac{d^{2} x}{d t^{2}}+2 \gamma \frac{d x}{d t}+\omega_{0}^{2} x=F \sin \omega_{0} t $$ Use Laplace transforms to find the displacement at a general time if the oscillator starts from rest at its equilibrium position. (a) Show that ultimately the oscillation has amplitude \(F /\left(2 \omega_{0} \gamma\right)\), with a phase lag of \(\pi / 2\) relative to the driving force per unit mass \(F\). (b) By differentiating the original equation, conclude that if \(x(t)\) is expanded as a power series in \(t\) for small \(t\), then the first non-vanishing term is \(F \omega_{0} t^{3} / 6\). Confirm this conclusion by expanding your explicit solution.

Solve the following first-order equations for the boundary conditions given: (a) \(y^{\prime}-(y / x)=1, \quad y(1)=-1 ;\) (b) \(y^{\prime}-y \tan x=1, \quad y(\pi / 4)=3\); (c) \(y^{\prime}-y^{2} / x^{2}=1 / 4, \quad y(1)=1\); (d) \(y^{\prime}-y^{2} / x^{2}=1 / 4, \quad y(1)=1 / 2\).

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