Question

For the following exercises, solve the differential equations. \(\frac{d y}{d x}=\sin ^{2} x\). The curve passes through point \((0,0)\).

Short answer

The solution is \( y = \frac{x}{2} - \frac{1}{4} \sin(2x) \).

Step-by-step solution

Separate Variables

The given differential equation is \( \frac{d y}{d x} = \sin^2 x \). To separate the variables, we write the equation as \( d y = \sin^2 x \, d x \).

Integrate Both Sides

Integrate both sides of the equation. On the left, integrate with respect to \( y \), and on the right, integrate \( \sin^2 x \) with respect to \( x \). Thus, we have \( \int d y = \int \sin^2 x \, d x \).

Solve for the Integral of sin-squared

To integrate \( \sin^2 x \), use the identity \( \sin^2 x = \frac{1 - \cos(2x)}{2} \). Substitute this into the integral, \( \int \sin^2 x \, dx = \int \frac{1 - \cos(2x)}{2} \, dx \).

Integrate Simplified Expression

Integrate \( \int \frac{1}{2} - \frac{1}{2} \cos(2x) \, dx \). This becomes \( \int \frac{1}{2} \, dx - \int \frac{1}{2} \cos(2x) \, dx \).

Calculate the Integrals

Calculate both integrals: \( \int \frac{1}{2} \, dx = \frac{x}{2} \) and \( \int \frac{1}{2} \cos(2x) \, dx = \frac{1}{4} \sin(2x) \). Thus, the integral is \( \frac{x}{2} - \frac{1}{4} \sin(2x) + C \).

Solve for y

Thus, \( y = \frac{x}{2} - \frac{1}{4} \sin(2x) + C \).

Determine the Constant C

Use the initial condition \( (0,0) \): substitute \( x = 0 \) and \( y = 0 \) into \( y = \frac{x}{2} - \frac{1}{4} \sin(2x) + C \). This gives \( 0 = \frac{0}{2} - \frac{1}{4} \sin(0) + C \), so \( C = 0 \).

Write the Final Solution

Substituting \( C = 0 \) into the equation gives the final solution as \( y = \frac{x}{2} - \frac{1}{4} \sin(2x) \).

Key concepts

Variable Separation

When faced with a differential equation such as \( \frac{dy}{dx} = \sin^2 x \), separating variables is often the first step in finding its solution. This method involves rearranging the equation to isolate each variable on opposite sides, making it easier to integrate.

In this case, we rewrite it so that all terms involving \( y \) are on one side of the equation, and all terms involving \( x \) are on the other. Thus, we write \( dy = \sin^2 x \, dx \).
This separation is crucial because it sets the stage for integrating both sides independently.

This step underscores the beauty and simplicity of variable separation, allowing us to tackle the problem through integration.

Integral Calculus

Once we have successfully separated the variables, the next step is to integrate both sides of the equation. This is where integral calculus comes into play.

On the left, we have \( \int dy \), which is straightforwardly \( y + C \), where \( C \) is the constant of integration. For the right side, \( \int \sin^2 x \, dx \), we need to apply a useful trigonometric identity: \( \sin^2 x = \frac{1 - \cos(2x)}{2} \). Substituting this into the integral simplifies it to \( \int \left( \frac{1}{2} - \frac{1}{2} \cos(2x) \right) dx \).

The integration of these simpler terms \( \frac{1}{2}x \) and \( -\frac{1}{4} \sin(2x) \) results in the expression \( \frac{x}{2} - \frac{1}{4} \sin(2x) + C \).

This process highlights the power and efficiency of integral calculus in solving differential equations, allowing us to find a general solution from a previously complex form.

Initial Condition

Every solution to a differential equation that involves an integration constant needs an initial condition to find a unique solution. This initial condition is given as a point through which the solution curve passes.

For the equation \( y = \frac{x}{2} - \frac{1}{4} \sin(2x) + C \), the curve passes through the point \((0,0)\). By substituting \( x = 0 \) and \( y = 0 \) into the equation, we can solve for \( C \). This results in \( 0 = 0 - 0 + C \), and therefore, \( C = 0 \).

Using this initial condition, we can refine the general solution to the specific solution \( y = \frac{x}{2} - \frac{1}{4} \sin(2x) \).

Initial conditions are vital in determining unique solutions that fit real-world constraints or specific problems. They help bridge the gap from a general formula to a specific, applicable one.