Question

Solve the given differential equation. $$ y^{\prime}+y^{2} \sin x=0 $$

Short answer

Question: Solve the differential equation \(y' + y^2 \sin x = 0\). Answer: The general solution for the given differential equation is \(y(x) = \frac{1}{-\cos x + \frac{1}{C} - 1}\).

Step-by-step solution

Rewrite the differential equation in separated form

To start solving the differential equation, first rewrite the equation \(y' + y^2 \sin x = 0\) by separating the variables. To do this, move the \(y^2 \sin x\) to the other side of the equation: $$ y' = -y^2 \sin x $$ Now, divide both sides by \(y^2\) and multiply both sides by \(\sin x\), resulting in the following separated form: $$ \frac{dy}{y^2} = -\sin x dx $$

Integrate both sides

Integrate both sides of the equation with respect to their respective variables: $$ \int \frac{dy}{y^2} = -\int \sin x dx $$ On the left side, the integral evaluates to: $$ -\frac{1}{y} $$ On the right side, the integral evaluates to: $$ \cos x + C_1 $$ Now, our equation stands as follows: $$ -\frac{1}{y} = \cos x + C_1 $$

Solve for y(x)

To find the general solution, y(x), we first need to isolate y from the equation. To do this, multiply both sides of the equation by \(-1\): $$ \frac{1}{y} = -\cos x - C_1 $$ Now, take the reciprocal of both sides: $$ y = \frac{1}{-\cos x - C_1} $$ To express the solution in terms of an arbitrary constant, \(C\), rewrite \(C_1\) as follows: $$ C_1 = \frac{1}{C} - 1 $$ Substituting this expression back into our general solution yields the final result: $$ y(x) = \frac{1}{-\cos x + \frac{1}{C} - 1} $$ This is the general solution for the given differential equation.

Key concepts

Separable Differential Equations

When we're faced with a differential equation, finding a way to solve it can often feel like a puzzle. For certain types of these puzzles, there's a handy strategy we can use, called separating variables. This strategy applies specifically to separable differential equations, which are equations where we can arrange all the terms involving the dependent variable (often 'y') on one side and all terms involving the independent variable (usually 'x') on the other side.

In our exercise, the original equation, \(y' + y^2 \sin x = 0\), is already whispering hints that it might be separable. The first step is to do some rearranging to isolate \(y'\). This gives us \(y' = -y^2 \sin x\), which is not quite separated yet. But with a little more shuffling—dividing through by \(y^2\) and multiplying by \(dx\)—we can split it into a pair of integrals, with \(y\) on one side and \(x\) on the other, like this: \[\int \frac{dy}{y^2} = -\int \sin x \, dx\]. This separation sets us up for success, transforming the differential equation into a form that's ready to be tackled with integration techniques.

Integration Techniques

Once we've managed to get our differential equation in a separable form, the next conquest is to pull out our integration techniques. These techniques are nothing short of magical mathematical tools that allow us to find the antiderivatives of functions.

In the given exercise, we encounter two integrals that we need to solve: \[\int \frac{dy}{y^2}\] and \[\int -\sin x \, dx\]. The left side is a straightforward power rule scenario, where the integral of \(1/y^2\) with respect to \(y\) can be looked at as \(y^{-2}\), leading to \( -1/y \). The right side is a basic trigonometric integral—everyone's favorite—where the integral of \(\sin(x)\) is \(\cos(x)\), and since we're integrating \( -\sin(x)\), we add the trigonometric twist to get \(\cos(x)\) plus an arbitrary constant, \(C_1\).

Unlocking the antiderivatives is like finding the missing pieces of our puzzle. However, the integration part is merely a step in this journey—you still need to complete the quest by solving for the dependent variable to get your final answer.

General Solution of Differential Equations

Stepping through the magical gateway of integration, we now seek to discover the general solution of the differential equation—a formula expressing all possible solutions. In essence, it's a treasure trove of functions that could possibly satisfy the original equation.

With the integration done, our exercise has guided us to an expression where \(y\) is alone on one side, signaling that we're close to stating the general solution. We maneuver out of the fractional inverse that we find ourselves in by taking the reciprocal of both sides. In doing so, we ensure that every move aligns \(y\) in its proper place, which finally gives us \(y(x) = 1/(-\cos x - C_1)\).

This form represents all the infinite possibilities of \(y\) in terms of \(x\), considering the various constants that could come into play. By expressing \(C_1\) in terms of another arbitrary constant \(C\), we consolidate this infinite range into a more standard form, effectively locking in our general solution. This form is not only essential for understanding the behavior of solutions under different initial conditions but is also a cornerstone in the study of differential equations.