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Chapter 17: Problem 41

The general solution of the differential equation \(\frac{d y}{d x}+y g^{\prime}(x)=\mathrm{g}(x) \cdot g^{\prime}(x)\) where \(g(x)\) is a given function of \(x\), is (A) \(g(x)+\log [1+y+g(x)]=C\) (B) \(g(x)+\log [1+y-g(x)]=C\) (C) \(g(x)-\log [1+y-g(x)]=C\) (D) None of these

Short Answer

Expert verified
(A) \(g(x) + \log [1+y+g(x)] = C\)

Step by step solution

01

Identify the Type of Differential Equation

The given differential equation is \( \frac{d y}{d x} + y g^{\prime}(x) = g(x) \cdot g^{\prime}(x) \). Its structure suggests it is a first-order linear differential equation, which can be solved using an integrating factor.
02

Determine the Integrating Factor

For a differential equation of the form \( \frac{dy}{dx} + P(x)y = Q(x) \), the integrating factor (I.F) is \( \mu(x) = e^{\int P(x) \, dx} \). Here, \( P(x) = g^{\prime}(x) \), thus the integrating factor is \( \mu(x) = e^{\int g^{\prime}(x) \, dx} = e^{g(x)} \).
03

Apply the Integrating Factor

Multiply the entire differential equation by the integrating factor \( e^{g(x)} \): \[ e^{g(x)} \frac{dy}{dx} + e^{g(x)} y g^{\prime}(x) = e^{g(x)} g(x) g^{\prime}(x) \].
04

Simplify to Find an Exact Derivative

Notice that the left-hand side is the derivative of \( e^{g(x)} y \) with respect to \( x \): \( \frac{d}{dx} \left( e^{g(x)} y \right) = e^{g(x)} g(x) g^{\prime}(x) \).
05

Integrate Both Sides

Integrate both sides with respect to \( x \): \[ \int \frac{d}{dx} (e^{g(x)} y) \, dx = \int e^{g(x)} g(x) g^{\prime}(x) \, dx \]. The left-hand side simplifies to \( e^{g(x)} y \).
06

Integrate the Right-Hand Side

The integral \( \int e^{g(x)} g(x) g^{\prime}(x) \, dx \) can be simplified by substitution. Let \( u = g(x) \), then \( du = g^{\prime}(x) \, dx \). The integral becomes \( \int u e^u \, du \), which equals \( e^u (u - 1) + C \), thus \( e^{g(x)}(g(x) - 1) + C \).
07

Solve for the General Solution

Equating both integrated sides, we get \( e^{g(x)} y = e^{g(x)} (g(x) - 1) + C \). Let's express \( y \) explicitly: \( y = g(x) - 1 + Ce^{-g(x)} \).
08

Choose the Correct Option Based on Structure

The form of the solution suggests manipulating terms to fit given options. Test equivalence by re-arranging and re-evaluating the solution structure against provided options.

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Key Concepts

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

integrating factor method
The integrating factor method is a powerful technique that simplifies solving first-order linear differential equations. We use it when faced with equations of the form \( \frac{dy}{dx} + P(x)y = Q(x) \). By leveraging an integrating factor, we transform the equation into a form that is easier to solve.

To find the integrating factor, \( \mu(x) \), we compute \( e^{\int P(x) \, dx} \). This factor helps in rewriting the differential equation so we can recognize it as a derivative of a product.

Let鈥檚 break it down step-by-step:
  • Identify \( P(x) \) in your equation. It is the coefficient of \( y \).
  • Compute the integrating factor: \( \mu(x) = e^{\int P(x) \, dx} \).
  • Multiply the entire differential equation by this integrating factor.
  • Recognize the left-hand side as the derivative of the product of the integrating factor and \( y \).
The simplicity comes from the transformation, making the equation ready to be integrated directly.
general solution of differential equations
When solving differential equations, finding the general solution is key. It represents the family of solutions that incorporates all possible specific solutions of the equation. A general solution comprises constants that can be adjusted to satisfy unique initial conditions, thereby giving us a specific solution.

For first-order linear differential equations, after applying the integrating factor method, the left and side becomes a derivative, making it easier to integrate.
  • Integrate both sides of the equation after applying the integrating factor.
  • The result on the left will typically present as a function multiplied by \( y \), which can be separately determined.
  • The constants involved in the integration become part of the solution expression.
This approach allows you to clearly express \( y \) in terms of \( x \) and arbitrary constants, revealing the general nature of the solution.
differential equations
Differential equations play a central role in various fields including physics, engineering, and finance, as they describe how quantities change. Specifically, a differential equation involves an unknown function and its derivatives.

The simplest form, a first-order linear differential equation (like the one in our exercise), includes only the first derivative of the unknown function. These equations have the form \( \frac{dy}{dx} + P(x)y = Q(x) \), where \( P(x) \) and \( Q(x) \) are functions of \( x \).
  • "First-order" signifies that it only involves the first derivative \( \frac{dy}{dx} \).
  • "Linear" means it is linear in terms of \( y \) and its derivative.
Understanding these principles gives insight into analyzing such equations and applying methods like the integrating factor method to find their solutions.

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Most popular questions from this chapter

Let the population of rabbits surviving at a time \(t\) be governed by the differential equation \(\frac{d p(t)}{d t}=\frac{1}{2} p(t)-200 .\) If initially \(p(0)=100\), then \(p(t)\) equals [2014] (A) \(400-300 e^{[/ 2}\) (B) \(300-200 e^{-t / 2}\) (C) \(600-500 e^{t / 2}\) (D) \(400-300 e^{-t / 2}\)

For a certain curve \(y=f(x)\) satisfying \(\frac{d^{2} y}{d x^{2}}=6 x-4\), \(f(x)\) has a local minimum value 5 when \(x=1\). (A) Equation of the curve is \(y=x^{3}-2 x^{2}+x+5\) (B) \(f(x)\) has a local maximum at \(x=\frac{1}{3}\) (C) Global maximum value of \(f(x)\) is 7 (D) Global minimum value of \(f(x)\) is 5

Passage 2 Some equations which are not exact can be made exact on multiplication by some suitable function known as an integrating factor. The equation $$ x d y-y d x=0 $$ which is not exact becomes so on multiplication by \(1 / y^{2}\), for then we $$ \frac{x}{y^{2}} d y-\frac{1}{y} d x=0 $$ which is easily seen to be exact. We can solve it either by re-arranging the terms and making them exact differential or by the method of exact equations. We now give some rules for finding integrating factors of differential equation $$ M d x+N d y=0 $$ to make it exact. I. If \(M x+N y \neq 0\) and the equation is homogeneous, then \(\frac{1}{M x+N y}\) is an I.F. II. If the equation \(M d x+N d y=0\) is not exact but is of the form \(f_{1}(x y) y d x+f_{2}(x y) x d y=0\), then \(\frac{1}{M x-N y}\) is an I.F., provided \(M x-N y \neq 0\) III. When \(\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{N}\) is a function of \(x\) alone, say \(f(x)\), then I.F. \(=e^{\int f(x) d x}\) IV. When \(\frac{\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}}{M}\) is a function of \(y\) alone, say \(f(y)\), then I.F. \(=e^{\int f(y) d y}\) The solution of the equation \((x y \sin x y+\cos x y) y d x+(x y \sin x y-\cos x y) x d y=0\) is (A) \(y \sec x y=c x\) (B) \(x \sec x y=c y\) (C) \(x \operatorname{cosec} x y=c y\) (D) None of these

The solution of the differential equation \(\frac{d y}{d x}=\frac{x+y}{x}\) satisfying the condition \(y(1)=1\) is \([2008]\) (A) \(y=\ln x+x\) (B) \(y=x \ln x+x^{2}\) (C) \(y=x e^{(x-1)}\) (D) \(y=x \ln x+x\)

The solution of the differential equation \((x-y)(2 d y-d x)=3 d x-5 d y\) is (A) \(2 x-y=\log (x-y+z)+c\) (B) \(2 x+y=\log (x-y+z)+c\) (C) \(2 y-x=\log (x-y+z)+c\) (D) None of these
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