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Chapter 24: Problem 62

The solution of the differential equation \(x \frac{d y}{d x}+2 y=x^{2}\) \((x \neq 0)\) with \(y(1)=1\), is: (a) \(y=\frac{4}{5} x^{3}+\frac{1}{5 x^{2}}\) (b) \(y=\frac{x^{3}}{5}+\frac{1}{5 x^{2}}\) (c) \(y=\frac{x^{2}}{4}+\frac{3}{4 x^{2}}\) (d) \(y=\frac{3}{4} x^{2}+\frac{1}{4 x^{2}}\)

Short Answer

Expert verified
The solution is (c) \(y=\frac{x^{2}}{4} + \frac{3}{4 x^{2}}\).

Step by step solution

01

Identify the type of differential equation

The given differential equation is \(x \frac{dy}{dx} + 2y = x^2\). This is a first-order linear differential equation, which can be rewritten in the standard form \(\frac{dy}{dx} + P(x)y = Q(x)\). By dividing through by \(x\), we get \(\frac{dy}{dx} + \frac{2}{x}y = x\).
02

Determine the integrating factor

To solve the equation, we need to find the integrating factor, \( I(x) \). The integrating factor for a first-order linear differential equation \( \frac{dy}{dx} + P(x)y = Q(x) \) is given by \( I(x) = e^{\int P(x) \, dx} \). Here, \( P(x) = \frac{2}{x} \). Thus, the integrating factor is \( I(x) = e^{\int \frac{2}{x} \, dx} = e^{2\ln|x|} = x^2 \).
03

Multiply through by the integrating factor

Multiply the entire differential equation by the integrating factor \(x^2\):\[x^2 \frac{dy}{dx} + 2xy = x^3\]
04

Rewrite as a derivative

Notice that the left-hand side of the equation is now in the form of a derivative: \( \frac{d}{dx}(x^2y) = x^3 \). This means the equation can be rewritten as:\[\frac{d}{dx}(x^2y) = x^3\]
05

Integrate both sides

Integrate both sides with respect to \( x \):\[\int \frac{d}{dx}(x^2y) \, dx = \int x^3 \, dx\]The left side simplifies to \(x^2 y\), and the right side integrates to \(\frac{x^4}{4} + C\). Thus,:\[ x^2 y = \frac{x^4}{4} + C \]
06

Solve for \(y\)

Divide both sides by \(x^2\) to solve for \(y\):\[y = \frac{x^4}{4x^2} + \frac{C}{x^2} = \frac{x^2}{4} + \frac{C}{x^2}\]
07

Apply the initial condition

Use the initial condition \( y(1) = 1 \) to find \( C \). Substitute \( x = 1 \) and \( y = 1 \) into the equation:\[1 = \frac{1^2}{4} + \frac{C}{1^2}\]This simplifies to \(1 = \frac{1}{4} + C\). Solving for \( C \), we find that:\[ C = 1 - \frac{1}{4} = \frac{3}{4} \]
08

Write the particular solution

Substitute \( C = \frac{3}{4} \) into the equation for \( y \):\[ y = \frac{x^2}{4} + \frac{3}{4x^2} \]

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Key Concepts

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

First-Order Linear Differential Equation
First-order linear differential equations are a fundamental part of calculus and differential equations. These equations involve functions and their first derivatives. They can be written in the standard form:
  • \( \frac{dy}{dx} + P(x)y = Q(x) \)
This form helps to identify and solve them more easily. In this exercise, the initial differential equation was \( x \frac{dy}{dx} + 2y = x^2 \). By dividing the entire equation by \( x \), it transforms into the standard form:
  • \( \frac{dy}{dx} + \frac{2}{x}y = x \)
Understanding how to manipulate the given equation into the standard form is crucial for applying further solution methods, such as the integrating factor technique.
Integrating Factor
The integrating factor is a powerful tool used to solve first-order linear differential equations. Once the equation is in its standard form \( \frac{dy}{dx} + P(x)y = Q(x) \), we find the integrating factor \( I(x) \). It is given by:
  • \( I(x) = e^{\int P(x) \, dx} \)
For the problem at hand, \( P(x) = \frac{2}{x} \). Therefore, the integrating factor becomes:
  • \( I(x) = e^{\int \frac{2}{x} \, dx} = e^{2\ln|x|} = x^2 \)
Multiplying the differential equation by this integrating factor turns the left-hand side into the derivative of a product, simplifying the integration process significantly, and allowing us to solve for \( y \) more easily.
Initial Condition
Initial conditions are essential in finding a particular solution to a differential equation. They specify a value of the solution at a given point, ensuring uniqueness of the solution. For this exercise, the initial condition provided is \( y(1) = 1 \).
After solving the general solution, \( y = \frac{x^2}{4} + \frac{C}{x^2} \), we substitute \( x = 1 \) and \( y = 1 \) to find \( C \).
  • Setting up the equation: \( 1 = \frac{1}{4} + C \)
  • Solving yields \( C = \frac{3}{4} \)
Inserting \( C = \frac{3}{4} \) back into the general solution gives the particular solution:
  • \( y = \frac{x^2}{4} + \frac{3}{4x^2} \)
This ensures that the solution satisfies the given initial condition and aligns with the specifics of the problem.

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Most popular questions from this chapter

Let \(f(x)=\left(\sin \left(\tan ^{-1} x\right)+\sin \left(\cot ^{-1} x\right)\right)^{2}-1,|x|>1\). If \(\frac{d y}{d x}=\frac{1}{2} \frac{d}{d x}\left(\sin ^{-1}(f(x))\right)\) and \(y(\sqrt{3})=\frac{\pi}{6}\), then \(y(-\sqrt{3})\) is equal to: (a) \(\frac{2 \pi}{3}\) (b) \(-\frac{\pi}{6}\) (c) \(\frac{5 \pi}{6}\) (d) \(\frac{\pi}{3}\)

The integrating factor of the differential equation \(\left(x^{2}-1\right) \frac{d y}{d x}+2 x y=x\) is (a) \(\frac{1}{x^{2}-1}\) (b) \(x^{2}-1\) (c) \(\frac{x^{2}-1}{x}\) (d) \(\frac{x}{x^{2}-1}\)

Solution of the differential equation \(y d x+\left(x+x^{2} y\right) d y=0\) is (a) \(\log y=C x\) (b) \(-\frac{1}{x y}+\log y=C\) (c) \(\frac{1}{x y}+\log y=C\) (d) \(-\frac{1}{x y}=C\)

If \(f(x)\) is a differentiable function in the interval \(((0, \infty)\) such that \(\mathrm{f}(\mathrm{a})=1\) and \(\lim _{\mathrm{t} \rightarrow \mathrm{x}} \frac{\mathrm{t}^{2} \mathrm{f}(\mathrm{x})-\mathrm{x}^{2} \mathrm{f}(\mathrm{t})}{\mathrm{t}-\mathrm{x}}=1\), for each \(\mathrm{x}>0\). then \(\mathrm{f}\left(\frac{3}{2}\right)\) is equal to: (a) \(\frac{23}{18}\) (b) \(\frac{13}{6}\) (c) \(\frac{25}{9}\) (d) \(\frac{31}{18}\)

Let \(y=y(x)\) be the solution of the differential equation, \(\frac{2+\sin x}{y+1} \cdot \frac{d y}{d x}=-\cos x, y>0, y(0)=1 .\) If \(y(\pi)=a\) and \(\frac{d y}{d x}\) at \(x=\pi\) is \(b\), then the ordered pair \((a, b)\) is equal to : (a) \(\left(2, \frac{3}{2}\right)\) (b) \((1,-1)\) (c) \((1,1)\) (d) \((2,1)\)
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