Problem 40 - Thomas Calculus

Question

Solve the initial value problems in Exercises \(37-40\) for \(y\) as a function of \(x .\) $$ \left(x^{2}+1\right)^{2} \frac{d y}{d x}=\sqrt{x^{2}+1}, \quad y(0)=1 $$

Step-by-Step Solution

Answer

The solution is \(y = -(x^2+1)^{-1/2} + 2\).

Step 1: Introduction to the Problem

We are given a differential equation \((x^2 + 1)^2 \frac{dy}{dx} = \sqrt{x^2 + 1}\) with an initial condition \(y(0) = 1\). We need to find the function \(y\) in terms of \(x\).

Step 2: Separate Variables

Separate the variables by rearranging the equation to isolate \(\frac{dy}{dx}\) on one side: \[\frac{dy}{dx} = \frac{\sqrt{x^2 + 1}}{(x^2 + 1)^2}.\]

Step 3: Integrate Both Sides

Integrate both sides with respect to their respective variables:\[\int dy = \int \frac{\sqrt{x^2 + 1}}{(x^2 + 1)^2} \, dx.\]This simplifies to:\[y = \int (x^2 + 1)^{-3/2} \, dx.\]

Step 4: Integration Using Substitution

Apply substitution method by setting \(u = x^2 + 1\) so that \(du = 2x \, dx\) and adjust the limits of the integration accordingly. This gives:\[y = \int u^{-3/2} \, \frac{du}{2}.\]

Step 5: Solve the Integral

Solve the integral:\[\frac{1}{2} \int u^{-3/2} \, du = \frac{1}{2} \cdot (-2)(x^2+1)^{-1/2} + C = -(x^2+1)^{-1/2} + C.\]Thus, the expression for \(y\) becomes:\[y = -(x^2+1)^{-1/2} + C.\]

Step 6: Apply the Initial Condition

Use the initial condition \(y(0) = 1\) to find the constant \(C\):\[1 = -\frac{1}{\sqrt{0^2+1}} + C \]\[1 = -1 + C \]\[C = 2.\]

Step 7: Final Solution

Replace \(C\) in the expression for \(y\):\[y = -(x^2+1)^{-1/2} + 2.\] This is the solution for the differential equation satisfying the initial condition.

Key Concepts

Initial Value Problem

An initial value problem in differential equations involves finding a function that satisfies a given differential equation and meets a specific initial condition. Here, we are tasked with solving a differential equation:

• The equation is \[ (x^2 + 1)^2 \frac{dy}{dx} = \sqrt{x^2 + 1} \\]
• The initial condition is given as \[ y(0) = 1. \\]

This means we need to find the function \(y(x)\) such that it satisfies both the differential equation and the initial condition at \(x = 0\). Initial conditions help determine the specific solution to a differential equation from the infinite set of potential solutions. This forms a complete profile of the function across its domain, ensuring a unique pathway we solve for.

Separation of Variables

Separation of variables is a technique where we rearrange a differential equation to isolate the dependent variable and its differential on one side and the independent variable and its differential on the other side. In this problem, we used this approach to manipulate the given differential equation:

• We started with \[ (x^2 + 1)^2 \frac{dy}{dx} = \sqrt{x^2 + 1} \\]
• By moving terms, we obtained \[ \frac{dy}{dx} = \frac{\sqrt{x^2 + 1}}{(x^2 + 1)^2} \\]

This equation is expressed such that \(dy\) and \(dx\) are "separated" from each other. Now, each side of the equation forms a part that can be individually integrated. This transformation is crucial as it allows us to solve the equation using integrations.

Integration Techniques

Once variables are separated, integration is used to solve the differential equation. Here, integration needs to be performed on both sides of the equation. We moved forward as follows:

• We integrated the left side simply as \[ \int dy = y + C \\]
• For the right side, \[ \int \frac{\sqrt{x^2 + 1}}{(x^2 + 1)^2} \, dx, \\]

we first simplified \[ (x^2+1)^{-3/2} \, dx, \\]which is integral using more advanced techniques. Each integration serves to take the differential equations from their differentiated forms back into an understandable and usable function form. This process allows us to determine the unknown function \(y(x)\) with respect to \(x\).

Substitution Method

Substitution is a common integration technique used to make complex integrands simpler and easier to evaluate. For this problem:

• The substitution was introduced as \(u = x^2 + 1\)
• By differentiating, we have \(du = 2x \, dx\), from which \( dx = \frac{du}{2x} \). However, since \( x\) was not present initially in our simplified integral, it simplifies solving.

Recasting the integral:

• This becomes a more straightforward integral \[ \int u^{-3/2} \, \frac{du}{2}, \\]

which is solvable using basic power rule reversals. This method helps us find antiderivatives that we can then use to solve the overall differential problem. It demonstrates how substitution can significantly simplify solving even complex looking differential equations.