Question

Use the method of reduction of order to find a second solution of the given differential equation. \(x y^{\prime \prime}-y^{\prime}+4 x^{3} y=0, \quad x>0 ; \quad y_{1}(x)=\sin x^{2}\)

Short answer

The expression for the second independent solution \(y_2(x)\) is: \(y_2(x) = \sin x^2 \cdot (\int \frac{C_1}{x^2 \sin x^2} dx + C_2)\), where \(C_1\) and \(C_2\) are constants of integration.

Step-by-step solution

Write expression for \(y_2(x)\) and derivatives

First, let's express the second solution as \(y_2(x)=y_1(x) \cdot v(x)\), with \(y_1(x)=\sin x^2\) already given. Next, we need to determine the derivatives \(y_2^{\prime}(x)\) and \(y_2^{\prime\prime}(x)\), which we will later substitute into the given ODE. Here are the expressions for \(y_2(x)\) and its derivatives: \(y_2(x) = \sin x^2 \cdot v(x)\) \(y_2^{\prime}(x) = \cos x^2 \cdot 2x \cdot v(x) + \sin x^2 \cdot v^{\prime}(x)\) \(y_2^{\prime\prime}(x) = \cos x^2 \cdot (-4x^2 \sin x^2)\cdot v(x) + 2 [\cos x^2 \cdot 2x \cdot v(x)]^{\prime} + \sin x^2 \cdot v^{\prime\prime}(x)\)

Substitute expressions for \(y_2(x)\) and derivatives into the given ODE

Now, let's substitute the expressions for \(y_2(x)\), \(y_2^{\prime}(x)\), and \(y_2^{\prime\prime}(x)\) into the given ODE: \(x[\cos x^2 \cdot (-4x^2 \sin x^2)\cdot v(x) + 2 [\cos x^2 \cdot 2x \cdot v(x)]^{\prime} + \sin x^2 \cdot v^{\prime\prime}(x)]-[2x \cos x^2 \cdot v(x)+\sin x^2 \cdot v^{\prime}(x)]+4x^3\sin x^2 \cdot v(x)=0\)

Simplify and derive first-order differential equation for \(v(x)\)

In this step, we want to simplify the equation and reduce it to a first-order differential equation for \(v(x)\): \(-4x^3\sin x^2 \cos x^2 \cdot v(x) + 4x^2 \cos^2 x^2 \cdot v^{\prime}(x) + x \sin x^2 \cdot v^{\prime\prime}(x) - 2x \cos x^2 \cdot v^{\prime}(x) = 0\) Now, let's factor out \(v(x)\) and its derivatives, so we can work with an equation for \(v(x)\) only: \(v^{\prime\prime}(x) - \frac{2x \cos x^2}{x\sin x^2} v^{\prime}(x) = 0\)

Solve the first-order differential equation for \(v(x)\)

Now, we solve the first-order differential equation for \(v(x)\): Let \(u(x) = v^{\prime}(x)\). Then the equation becomes: \(u^{\prime}(x) - \frac{2x \cos x^2}{x\sin x^2} u(x) = 0\) This equation is of first-order linear form, and we can use the integrating factor method to solve it. Let \(IF(x) = e^{\int \frac{-2x \cos x^2}{x\sin x^2} dx}\) be the integrating factor. This integral is hard to evaluate, but we can recognize that the differential of \(x^2\sin x^2\) is precisely the integrand: \(d(x^2\sin x^2)=-2x\cos x^2 dx\). Therefore, the integral evaluates to: \(IF(x) = e^{\ln(x^2\sin x^2)} = x^2\sin x^2\) Now, let's multiply the equation with the integrating factor: \(x^2 \sin x^2 \cdot u^{\prime}(x) - 2x \cos x^2 \cdot u(x) = 0\) The left-hand side is now the derivative of \(u(x) \cdot x^2 \sin x^2\): \( \frac{d}{dx}(u(x) \cdot x^2 \sin x^2) = 0\) Now, integrate both sides of the equation: \(u(x) \cdot x^2 \sin x^2 = C_1\) Next, remember that \(u(x)=v^{\prime}(x)\), so divide by \(x^2 \sin x^2\): \(v^{\prime}(x) = \frac{C_1}{x^2 \sin x^2}\) Integrate this equation to find \(v(x)\): \(v(x)= \int \frac{C_1}{x^2 \sin x^2} dx + C_2\) Unfortunately, this integral does not have an elementary closed form, so we must express the solution in terms of an integral: \(v(x) = \frac{C_1}{x^2 \sin x^2} dx + C_2\) Finally, we obtain the second independent solution by multiplying \(v(x)\) by \(y_1(x)\): \(y_2(x) = \sin x^2 \cdot (\int \frac{C_1}{x^2 \sin x^2} dx + C_2)\) Although the answer is expressed in an integral form, it represents the second independent solution to the given ODE.

Key concepts

Second Solution

In the world of differential equations, finding a second solution can often be a puzzle, especially when you already have the first solution. Suppose you have a differential equation along with one solution. In that case, you can apply a technique called Reduction of Order to discover a second, linearly independent solution. Reduction of Order is like having a compass that guides you to another path after you've already discovered the first. This technique combines a known solution with multiplication by another unknown function. In our case, it means expressing the second solution as:

• \( y_2(x) = y_1(x) \cdot v(x) \)

Here, \( y_1(x) = \sin x^2 \) is our given first solution, and \( v(x) \) is the unknown function we want to find. The beauty of this method is that it allows you to transform a complicated second-order problem into a more manageable form.

Differential Equation

A differential equation is like a mathematical equation whispering instructions about a function and its derivatives. Specifically, it tells us how changes in one quantity affect others when describing real-world phenomena. In our original problem, the differential equation is:

• \( x y^{\prime \prime} - y^{\prime} + 4x^3 y = 0 \)

This equation involves second derivatives, derivatives, and the function itself: a classic second-order differential equation. The beauty of such equations lies in their ability to model everything from oscillations in physics to growth rates in biology. But solving them requires savvy techniques like Reduction of Order, allowing us to untangle complex mathematical webs. As you step through a differential equation, remember that it controls how stringent or lenient a function must act to satisfy it across various boundaries and initial conditions.

First-Order Differential Equation

Taking a second-order differential equation and simplifying it to a first-order one is like turning a challenging puzzle into a simpler one. In step 3 of our solution, we ended up with a much simpler equation:

• \( v^{\prime\prime}(x) - \frac{2x \cos x^2}{x\sin x^2} v^{\prime}(x) = 0 \)

This step is crucial because first-order differential equations are generally easier to solve than their second-order siblings. We reduced a complex task by focusing on derivatives of \( v(x) \), our unknown function. These types of equations involve only the first derivative of the unknown function and are instrumental in many scenarios where interactions are straightforward. Everything from modeling velocity to predicting decay rates can be boiled down to first-order differential equations, making our life more straightforward in the world of math.

Integration Factor

The Integration Factor is a magical key to unlock linear first-order differential equations. We use it to transform equations into a form that makes them easy to solve, effectively easing our journey. After rewriting our main equation with \( u(x) = v^{\prime}(x) \), we recognize it fits the mold for using an integration factor. Here's the main step:
Calculate:

• Integration Factor (\( IF(x) \)): \( e^{\int \frac{-2x \cos x^2}{x\sin x^2} dx} \)

This seemingly cryptic exponential expression becomes a practical handrail, allowing us to maneuver through the equation smoothly. It simplifies the problem by converting the product of the integration factor with the derivative into the derivative of a product. This technique allows us to integrate easily and find the solution to our first-order differential equation quickly. By leveraging natural logs and recognizing derivative relations, the integration factor harmonizes complexities, making our path towards finding \( v(x) \) clear and direct. Each step of applying it unwraps layers of mathematical complexity, reflecting its critical role in solving linear differential equations.