Math Problem Statement

Eliminate arbitrary constants from the following equations: (a) ln y = C1 e^x + C2 e^{-x}, (b) y^{-3} = x^3(3e^x + C), (c) y = C1 e^{2x} cos 3x + C2 e^{2x} sin 3x, (d) y = C1 x^2 + C2 e^{2x}.

Solution

The problem asks to eliminate arbitrary constants from the following given equations. Let's address each part one by one:

a) lny=C1ex+C2ex\ln y = C_1 e^x + C_2 e^{-x}

To eliminate the arbitrary constants C1C_1 and C2C_2, we need to differentiate the equation multiple times until these constants disappear.

  1. First, exponentiate both sides to remove the logarithm: y=eC1ex+C2exy = e^{C_1 e^x + C_2 e^{-x}} However, differentiating directly from the original equation might be more efficient.

  2. Differentiate both sides with respect to xx: 1ydydx=C1exC2ex\frac{1}{y} \frac{dy}{dx} = C_1 e^x - C_2 e^{-x} Multiply both sides by yy: dydx=y(C1exC2ex)\frac{dy}{dx} = y(C_1 e^x - C_2 e^{-x})

  3. Differentiate again: d2ydx2=dydx(C1exC2ex)+y(C1ex+C2ex)\frac{d^2 y}{dx^2} = \frac{dy}{dx}(C_1 e^x - C_2 e^{-x}) + y(C_1 e^x + C_2 e^{-x}) This eliminates the constants, leading to a differential equation in yy.

b) y3=x3(3ex+C)y^{-3} = x^3(3e^x + C)

The strategy is to differentiate the equation to eliminate the constant CC.

  1. Differentiate both sides with respect to xx: 3y4dydx=3x2ex-3y^{-4} \frac{dy}{dx} = 3x^2 e^x

  2. Simplify the equation and solve for dydx\frac{dy}{dx}: dydx=x2exy43\frac{dy}{dx} = -\frac{x^2 e^x y^4}{3}

At this point, you have a differential equation without the constant CC.

c) y=C1e2xcos3x+C2e2xsin3xy = C_1 e^{2x} \cos 3x + C_2 e^{2x} \sin 3x

This is a second-order linear combination of sine and cosine. To eliminate C1C_1 and C2C_2, we differentiate twice.

  1. First derivative: dydx=C1(2e2xcos3x3e2xsin3x)+C2(2e2xsin3x+3e2xcos3x)\frac{dy}{dx} = C_1(2e^{2x} \cos 3x - 3e^{2x} \sin 3x) + C_2(2e^{2x} \sin 3x + 3e^{2x} \cos 3x)

  2. Second derivative: d2ydx2=C1(5e2xcos3x12e2xsin3x)+C2(5e2xsin3x+12e2xcos3x)\frac{d^2y}{dx^2} = C_1(-5e^{2x} \cos 3x - 12e^{2x} \sin 3x) + C_2(-5e^{2x} \sin 3x + 12e^{2x} \cos 3x)

You can now write the second derivative in terms of yy: d2ydx24dydx+13y=0\frac{d^2y}{dx^2} - 4 \frac{dy}{dx} + 13y = 0 This is the required differential equation without arbitrary constants.

d) y=C1x2+C2e2xy = C_1 x^2 + C_2 e^{2x}

To eliminate C1C_1 and C2C_2, differentiate the equation twice:

  1. First derivative: dydx=2C1x+2C2e2x\frac{dy}{dx} = 2C_1 x + 2C_2 e^{2x}

  2. Second derivative: d2ydx2=2C1+4C2e2x\frac{d^2y}{dx^2} = 2C_1 + 4C_2 e^{2x}

Now you have a system of equations involving y,dydx,y, \frac{dy}{dx}, and d2ydx2\frac{d^2y}{dx^2}, allowing you to eliminate C1C_1 and C2C_2.


These are the differential equations derived from the original equations after eliminating the arbitrary constants.

Do you need further details on any part of the process, or do you have specific questions about one of the steps?


Related Questions:

  1. What are the general methods for eliminating constants from an equation?
  2. How do first and second derivatives help in eliminating arbitrary constants?
  3. What type of differential equation is obtained after eliminating constants in part c?
  4. Can constants always be eliminated using differentiation in any type of function?
  5. How does the initial condition influence the solution to these types of problems?

Tip:

When eliminating constants, it's helpful to look for patterns in differentiation that will simplify terms involving constants, aiming to reduce the equation to a form that no longer contains them.

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Math Problem Analysis

Mathematical Concepts

Differentiation
Elimination of constants
Exponential functions
Logarithmic functions
Trigonometric functions

Formulas

Derivative of e^x: d/dx e^x = e^x
Derivative of ln y: d/dx ln y = 1/y * dy/dx
Product and chain rule for differentiation

Theorems

Differentiation rules
Elimination of arbitrary constants through differentiation

Suitable Grade Level

College level calculus