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Chapter 2: Problem 15

Find the solution of the initial value problem. $$ \left(x^{2}+1\right) y^{\prime}+3 x y=6 x, \quad y(0)=-1 $$

Short Answer

Expert verified
The solution is \(y = 2 - \frac{3}{(x^2 + 1)^{\frac{3/2}}}\).

Step by step solution

01

Rewrite the differential equation

The given differential equation is \((x^2 + 1)y' + 3xy = 6x\), and we want to rewrite it in standard linear form. This is \(y' + P(x)y = Q(x)\). Divide every term by \((x^2 + 1)\) to get: \(y' + \frac{3x}{x^2 + 1}y = \frac{6x}{x^2 + 1}\).
02

Identify the integrating factor

The form of integrating factor \(\mu(x)\) for a linear first-order differential equation \(y' + P(x)y = Q(x)\) is \(\mu(x) = e^{\int P(x) \, dx}\). In this problem, \(P(x) = \frac{3x}{x^2 + 1}\). Integrate \(P(x)\):\[\int \frac{3x}{x^2 + 1} \, dx = \frac{3}{2}\ln(x^2 + 1)\]. Thus, the integrating factor is \(\mu(x) = e^{\frac{3}{2}\ln(x^2 + 1)} = (x^2 + 1)^{\frac{3}{2}}\).
03

Multiply through by the integrating factor

Multiply the entire differential equation by the integrating factor \((x^2 + 1)^{\frac{3}{2}}\): \((x^2 + 1)^{\frac{3}{2}}y' + \frac{3x}{x^2 + 1}(x^2 + 1)^{\frac{3}{2}} y = \frac{6x}{x^2 + 1}(x^2 + 1)^{\frac{3}{2}}\), which simplifies to:\((x^2 + 1)^{\frac{3}{2}}y' + 3x(x^2 + 1)^{\frac{1}{2}} y = 6x (x^2 + 1)^{\frac{1}{2}}\).
04

Simplify to an exact differential

Recognize that the left-hand side is a derivative of a product: \(\frac{d}{dx}((x^2 + 1)^{\frac{3}{2}}y)\). Hence, the equation becomes: \[\frac{d}{dx}((x^2 + 1)^{\frac{3}{2}}y) = 6x(x^2 + 1)^{\frac{1}{2}}\].
05

Integrate both sides

Integrate both sides of the equation with respect to \(x\):\[(x^2 + 1)^{\frac{3}{2}}y = \int 6x(x^2 + 1)^{\frac{1}{2}} \, dx\]. To evaluate the right side, use the substitution \(u = x^2 + 1\), so \(du = 2x \, dx\), transforming the integral to: \[3\int u^{\frac{1}{2}} \, du = 3\cdot\frac{2}{3}u^{\frac{3}{2}} = 2(x^2 + 1)^{\frac{3}{2}} + C\]. Thus, \((x^2 + 1)^{\frac{3}{2}}y = 2(x^2 + 1)^{\frac{3}{2}} + C\).
06

Solve for y

We can now solve for \(y\) by dividing both sides by \((x^2 + 1)^{\frac{3}{2}}\):\[y = 2 + \frac{C}{(x^2 + 1)^{\frac{3}{2}}}\].
07

Apply the initial condition

Use the initial condition \(y(0) = -1\) to determine the constant \(C\). At \(x = 0\), \(y = -1\), so:\[-1 = 2 + \frac{C}{1}\]. Solving gives: \(C = -3\).
08

Write the final solution

Substitute \(C = -3\) back into the general solution to get the specific solution that satisfies the initial condition: \[y = 2 - \frac{3}{(x^2 + 1)^{\frac{3/2}}}\].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Initial Value Problem
An initial value problem in differential equations involves finding a function that satisfies both a differential equation and a starting condition. In our case, the differential equation is
  • \((x^2 + 1)y' + 3xy = 6x\)
The starting condition or initial condition is given as \(y(0) = -1\). This implies that when \(x = 0\), the value of \(y\) must be \(-1\). The solution involves not just finding a general solution of the differential equation, but also determining the specific solution that meets this condition.
Initial value problems are widely used in physical sciences and engineering, as they can model real-world systems where initial conditions are known.
Integrating Factor
The integrating factor technique is essential for solving linear first-order differential equations. It helps in rewriting the differential equation into an exact form, which can be integrated easily. For a linear equation of the form
  • \(y' + P(x)y = Q(x)\),
the integrating factor \(\mu(x)\) is given by:\[\mu(x) = e^{\int P(x) \, dx}.\]
In our example, \(P(x) = \frac{3x}{x^2 + 1}\), so \( \mu(x) \) becomes
  • \((x^2 + 1)^{\frac{3}{2}}\).
Multiplying through the differential equation by this integrating factor makes the left side an exact derivative, simplifying the integration process substantially.
Linear First-Order Differential Equation
This type of differential equation involves derivatives of a function and the function itself, without involving higher-order derivatives or non-linear terms. The standard form for a linear first-order differential equation is as follows:\[y' + P(x)y = Q(x)\].
In the exercise, the differential equation presented was rearranged to this standard linear form, allowing us to use a systematic approach for solving it. These equations are fundamental in modeling processes such as population growth, radioactive decay, and many engineering problems, due to their simplicity and linearity, which imply that solutions can be directly added and scaled.
Exact Differential
The term 'exact differential' in the context of differential equations refers to a differential equation that can be derived from a simple derivative of a product of functions. Once the equation is multiplied by the integrating factor, it becomes:
  • \(\frac{d}{dx}((x^2 + 1)^{\frac{3}{2}}y) = 6x(x^2 + 1)^{\frac{1}{2}}\).
The left side is the derivative of a single product \((x^2 + 1)^{\frac{3}{2}}y\), making it an exact form.
This property greatly simplifies finding solutions, as you can directly integrate both sides, leading most often to a straightforward computation, solving the equation with much less room for error.

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Most popular questions from this chapter

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