Problem 15 - Differential Equations and Linear Algebra

Question

Solve the given initial-value problem. $$y^{\prime}=y^{3} \sin x, \quad y(0)=0$$

Step-by-Step Solution

Answer

The solution to the initial value problem \(y^{\prime}=y^{3} \sin x, y(0)=0\) is \(y(x) = -\frac{1}{\sqrt{2 \cos x - 2}}\).

Step 1: Rewrite the ODE into a separable form

We rewrite the given ODE as follows: \[ \frac{dy}{dx} = y^3 \sin x \] Now, we separate the variables by dividing both sides by \(y^3\): \[ \frac{1}{y^3} \frac{dy}{dx} = \sin x \]

Step 2: Integrate both sides

Now, we integrate both sides with respect to x. On the left side, we have \(y^3\) in the denominator, so we integrate \(\frac{1}{y^3}\) with respect to y. On the right side, we simply integrate \(\sin x\). \[ \int \frac{1}{y^3} \frac{dy}{dx} dx = \int \sin x dx \] The left side simplifies to: \[ \int \frac{1}{y^3} dy = \int \sin x dx \] Now, we can integrate both sides: \[ -\frac{1}{2y^2} = -\cos x + C_1 \]

Step 3: Solve for y

Now, we want to isolate y. We begin by adding \(\frac{1}{2y^2}\) to both sides and then multiplying both sides by -2: \[ \frac{1}{y^2} = 2 \cos x - 2C_1 \] Next, we flip both sides: \[ y^2 = \frac{1}{2 \cos x - 2C_1} \] Now, we take the square root of both sides to isolate y: \[ y(x) = \pm \frac{1}{\sqrt{2 \cos x - 2C_1}} \]

Step 4: Apply the initial condition

Now, we need to apply the initial condition \(y(0)=0\) to find the specific solution. Plug in \(x=0\) and \(y=0\) into the general solution: \[ 0 = \pm \frac{1}{\sqrt{2 \cos 0 - 2C_1}} \] This implies \(2 \cos 0 = 2C_1\), so \(C_1 = 1\).

Step 5: Write the final solution

Plug in the found value for the constant to obtain the exact solution: \[ y(x) = \pm \frac{1}{\sqrt{2 \cos x - 2}} \] Since the initial condition is \(y(0) = 0\), the positive square root does not satisfy the initial condition, so we take the negative square root: \[ y(x) = -\frac{1}{\sqrt{2 \cos x - 2}} \] This is the solution to the initial value problem.

Key Concepts

Separable Differential Equations

Separable differential equations are a specific type of differential equations where you can rewrite the equation in such a way that all terms involving one variable are on one side of the equation, and all terms involving the other variable are on the opposite side. This process is known as 'separation of variables'.

For example, consider the differential equation given: - \( \frac{dy}{dx} = y^3 \sin x \).
To make it separable, we can rewrite this as: - \( \frac{1}{y^3} \frac{dy}{dx} = \sin x \).
This is now in a form where we can integrate each side separately. Here is how it works:

• Move all \( y \) terms with \( dy \) to one side of the equation.
• Move all \( x \) terms with \( dx \) to the other side.

This straightforward yet powerful method allows us to solve certain types of differential equations that might otherwise be more complex.

Initial Value Problem

An initial value problem (IVP) not only seeks a function that satisfies a given differential equation but also meets specific conditions at a particular point. The condition is usually provided as a point on the function, like in the form \( y(x_0) = y_0 \).

In our example, the problem is:

• Differential equation: \( y' = y^3 \sin x \)
• Initial condition: \( y(0) = 0 \)

The initial condition essentially acts as a 'starting point' for the function and is critical in determining the specific solution among many possible general solutions. It's like providing a single clue from which the entire pathway of the solution can be traced uniquely.

Integration Techniques

Integration is the cornerstone technique used in solving separable differential equations. When solving these equations, once variables are separated, we integrate both sides of the equation—one with respect to \( y \) (using \( dy \)) and the other with respect to \( x \) (using \( dx \)).

For example, in solving \( \frac{1}{y^3} \frac{dy}{dx} = \sin x \), you perform these steps:

• Integrate \( \int \frac{1}{y^3} \, dy \)
• Integrate \( \int \sin x \, dx \)

The integration results in two expressions for each variable. The integral of \( \frac{1}{y^3} \, dy \) is \( -\frac{1}{2y^2} \). For the integral of \( \sin x \, dx \), it results in \( -\cos x \).

Integrating can be tricky, requiring various methods like substitution or partial fractions, depending on the complexity of the expression. But in separable differential equations, it often boils down to solving one-variable polynomials or trigonometric functions that are directly integrable without additional techniques.