Math Problem Statement

By the elimination of the constants 𝐴 and 𝐡 obtain the differential equation of which π‘₯𝑦 = 𝐴𝑒π‘₯ + 𝐡𝑒 βˆ’π‘₯ + π‘₯ 2 is the solution.

Solution

To derive the differential equation by eliminating the constants AA and BB, we are given the equation:

xy=Aex+Beβˆ’x+x2xy = A e^x + B e^{-x} + x^2

Step 1: Differentiate the given equation with respect to xx

We'll first differentiate the equation with respect to xx to eliminate the constants AA and BB. Apply the product rule for the term xyxy.

ddx(xy)=ddx(Aex+Beβˆ’x+x2)\frac{d}{dx}(xy) = \frac{d}{dx}(A e^x + B e^{-x} + x^2)

On the left-hand side: ddx(xy)=y+xdydx\frac{d}{dx}(xy) = y + x \frac{dy}{dx}

On the right-hand side: ddx(Aex+Beβˆ’x+x2)=Aexβˆ’Beβˆ’x+2x\frac{d}{dx}(A e^x + B e^{-x} + x^2) = A e^x - B e^{-x} + 2x

So, after differentiation, we get: y+xdydx=Aexβˆ’Beβˆ’x+2xy + x \frac{dy}{dx} = A e^x - B e^{-x} + 2x

Step 2: Differentiate again to eliminate constants AA and BB

We differentiate the result again to eliminate the constants AA and BB.

Differentiate the left-hand side: ddx(y+xdydx)=dydx+(dydx+xd2ydx2)\frac{d}{dx}\left(y + x \frac{dy}{dx}\right) = \frac{dy}{dx} + \left(\frac{dy}{dx} + x \frac{d^2y}{dx^2}\right) Simplifying: 2dydx+xd2ydx22 \frac{dy}{dx} + x \frac{d^2y}{dx^2}

Differentiate the right-hand side: ddx(Aexβˆ’Beβˆ’x+2x)=Aex+Beβˆ’x+2\frac{d}{dx}(A e^x - B e^{-x} + 2x) = A e^x + B e^{-x} + 2

Now, equate both sides: 2dydx+xd2ydx2=Aex+Beβˆ’x+22 \frac{dy}{dx} + x \frac{d^2y}{dx^2} = A e^x + B e^{-x} + 2

Step 3: Eliminate Aex+Beβˆ’xA e^x + B e^{-x}

From the original equation xy=Aex+Beβˆ’x+x2xy = A e^x + B e^{-x} + x^2, solve for Aex+Beβˆ’xA e^x + B e^{-x}:

Aex+Beβˆ’x=xyβˆ’x2A e^x + B e^{-x} = xy - x^2

Substitute this into the second derivative equation: 2dydx+xd2ydx2=(xyβˆ’x2)+22 \frac{dy}{dx} + x \frac{d^2y}{dx^2} = (xy - x^2) + 2

Simplifying: 2dydx+xd2ydx2=xyβˆ’x2+22 \frac{dy}{dx} + x \frac{d^2y}{dx^2} = xy - x^2 + 2

This is the required differential equation.

Summary

The differential equation of which xy=Aex+Beβˆ’x+x2xy = A e^x + B e^{-x} + x^2 is the solution is:

2dydx+xd2ydx2=xyβˆ’x2+22 \frac{dy}{dx} + x \frac{d^2y}{dx^2} = xy - x^2 + 2

Would you like further clarification on any step or details?


Here are 5 related questions:

  1. What is the role of differentiation in eliminating constants from a function?
  2. How does the product rule apply in solving differential equations?
  3. Why are constants like AA and BB significant in differential equations?
  4. How does a second derivative affect the behavior of a differential equation?
  5. Can the method used here be applied to other types of equations involving exponentials?

Tip: When solving differential equations, always keep track of your constants, as they can often be eliminated through differentiation steps.

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Exponential Functions
Product Rule

Formulas

xy = A e^x + B e^{-x} + x^2
First derivative: d/dx(xy) = y + x dy/dx
Second derivative: d/dx(y + x dy/dx) = 2 dy/dx + x d^2y/dx^2

Theorems

Product Rule
Exponential Function Derivatives

Suitable Grade Level

Undergraduate Mathematics