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Chapter 24: Problem 63

Let \(y=y(x)\) be the solution of the differential equation, \(\left(x^{2}+1\right)^{2} \frac{d y}{d x}+2 x\left(x^{2}+1\right) y=1\) such that \(y(0)=0\). If \(\sqrt{a}\) \(y(1)=\frac{\pi}{32}\), then the value of ' \(a\) ' is: (a) \(\frac{1}{4}\) (b) \(\frac{1}{2}\) (c) 1 (d) \(\frac{1}{16}\)

Short Answer

Expert verified
The value of 'a' is 1.

Step by step solution

01

Recognize Integrable Form

Identify that the given differential equation is in the form \[ (x^2 + 1)^2 \frac{dy}{dx} + 2x(x^2 + 1)y = 1, \]which resembles a first-order linear differential equation: \[ P(x)y + Q(x)\frac{dy}{dx} = R(x), \]indicating that you should use an integrating factor for its solution.
02

Calculate the Integrating Factor

To solve the differential equation, compute the integrating factor, which is given by: \[ \mu(x) = e^{\int P(x) \, dx}. \] Here, \( P(x) = \frac{2x}{(x^2 + 1)^2} \), hence: \[ \int P(x) \, dx = \int \frac{2x}{(x^2 + 1)^2} \, dx. \] Using substitution \( u = x^2 + 1 \), \( du = 2x \, dx \), the integral becomes \[ \int \frac{1}{u^2} \, du = -\frac{1}{u}, \]which gives the integrating factor: \[ \mu(x) = e^{-\frac{1}{x^2+1}}. \]
03

Simplify Using Integrating Factor

Multiply both sides of the differential equation by \( \mu(x) \) to simplify: \[ \mu(x) \cdot (x^2 + 1)^2 \frac{dy}{dx} + \mu(x) \, 2x(x^2 + 1)y = \mu(x). \]This turns the left-hand side into the derivative of \( \left[y \cdot (e^{-\frac{1}{x^2+1}} \cdot (x^2+1))\right] \), allowing us to integrate both sides.
04

Integrate Both Sides

Integrate the equation: \[ \int d[y \cdot (e^{-\frac{1}{x^2+1}} \cdot (x^2 +1))] = \int e^{-\frac{1}{x^2+1}} \, dx. \]Perform integration on the right-hand side to solve for \( y(x) \), using the boundary condition \( y(0) = 0 \) to find constants if necessary.
05

Apply Boundary Conditions

The integration gives an explicit formula \( y(x) \) that includes a constant from the integration. Apply the condition \( y(0) = 0 \) to compute this constant and finalize the function.
06

Determine Specific Value for x = 1

Use \( y(x) \) to calculate \( y(1) \), assuming that \( \sqrt{a} \cdot y(1) = \frac{\pi}{32} \), which requires us to deduce the value of ' \( a \) ' to satisfy the equation involving \( y(1) = \frac{\pi}{32} \).

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Key Concepts

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

Integrating Factor
When tackling a first-order linear differential equation, the integrating factor can be a useful tool. It's like finding the magic multiplier that makes solving the equation straightforward. For any equation in the form \( P(x)y + Q(x)\frac{dy}{dx} = R(x) \), the integrating factor is calculated as \( \mu(x) = e^{\int P(x) \, dx} \).
Imagine trying to integrate both sides of an equation, but the left side seems stubborn. The integrating factor makes the left side "behave" by transforming it into a form that's easy to integrate.
  • Calculate \( \mu(x) \).
  • Multiply every term in the equation by \( \mu(x) \).
  • Simplify to turn the left side into a simple derivative.
It's a systematic method that essentially "prepares" the equation for direct integration, helping you reach the solution without hassle.
First-Order Linear Differential Equation
First-order linear differential equations come up often in mathematical modeling. They involve derivatives and are the simplest form among differential equations.
In our problem, the equation was identified as \( (x^2 + 1)^2 \frac{dy}{dx} + 2x(x^2 + 1)y = 1 \). It fits the general form \( P(x)y + Q(x)\frac{dy}{dx} = R(x) \). This kind of equation usually involves one independent variable and its first derivative.
  • "First-order" indicates the highest derivative is the first derivative, \( \frac{dy}{dx} \).
  • "Linear" means that both the function and its derivative appear to the first power.
The beauty of first-order differential equations is their predictable nature - they can elegantly model phenomena where growth or decay processes are linear. This makes them widely useful in fields like physics and economics.
Initial Value Problem
An initial value problem is a differential equation along with a condition that specifies the value of the unknown function at a given point. This point acts like a starting marker that guides the trajectory of the solution curve.
In this exercise, we had the equation with a condition: \( y(0)=0 \). This means that when \( x = 0 \), the value of \( y \) must be 0. Such a condition helps to pin down the constant involved in the equation after integration.
  • Initial conditions provide context, ensuring the solution is unique and fits within an intended scenario.
  • Without these, you might find an infinite number of solutions.
By applying the initial condition, you can solve the equation and find specific constants, leading to a particular solution that aligns perfectly with the given scenario.

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Most popular questions from this chapter

The differential equation whose solution is \(A x^{2}+B y^{2}=1\) where \(\mathrm{A}\) and \(\mathrm{B}\) are arbitrary constants is of (a) second order and second degree (b) first order and second degree (c) first order and first degree (d) second order and first degree

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) \(-\frac{1}{x y}=C\)

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) \(\frac{\pi}{3}\)

If \(\mathrm{y}=\mathrm{y}(\mathrm{x})\) is the solution of the differential equation \(\frac{\mathrm{dy}}{\mathrm{dx}}=(\tan x-y) \sec ^{2} x, \mathrm{x} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\), such that \(\mathrm{y}(0)=0\) then \(y\left(-\frac{\pi}{4}\right)\) is equal to: (a) \(\mathrm{e}-2\) (b) \(\frac{1}{2}-e\) (c) \(2+\frac{1}{e}\) (d) \(\frac{1}{e}-2 \mid\)

Consider the differential equation: \(\frac{d y}{d x}=\frac{y^{3}}{2\left(x y^{2}-x^{2}\right)}\) Statement-1: The substitution \(z=y^{2}\) transforms the above equation into a first order homogenous differential equation. Statement-2: The solution of this differential equation is \(y^{2} e^{-y^{2} / x}=C\) (a) Both statements are false. (b) Statement- 1 is true and statement- 2 is false. (c) Statement- 1 is false and statement- 2 is true. (d) Both statements are true.
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