Math Problem Statement

Determine the differential equation whose solution is given by: 1. (y - 5)^2 = cx, 2. y = c1 + c2x + c3x^2 + c4x^3, 3. cx^2 + y^2 = 4, 4. y = c1e^(-x) + c2e^(2x) + c3e^(x).

Solution

The exercise you've shared is focused on deriving the differential equations by eliminating the arbitrary constants present in the given solutions. Let's go through each problem step by step to determine the corresponding differential equation.

1. (y5)2=cx(y - 5)^2 = cx

  • Differentiate both sides with respect to xx: 2(y5)dydx=c2(y - 5) \frac{dy}{dx} = c Rearranging this, we get: dydx=c2(y5)\frac{dy}{dx} = \frac{c}{2(y - 5)} The constant cc can be eliminated using the original equation (y5)2=cx(y - 5)^2 = cx, giving: c=(y5)2xc = \frac{(y - 5)^2}{x} Substituting this back into the derivative: dydx=(y5)2x\frac{dy}{dx} = \frac{(y - 5)}{2x} Multiplying both sides by 2x2x, we get the final differential equation: 2xdydx=y52x \frac{dy}{dx} = y - 5

2. y=c1+c2x+c3x2+c4x3y = c_1 + c_2 x + c_3 x^2 + c_4 x^3

  • Since there are 4 arbitrary constants, we'll need to differentiate multiple times to eliminate them.
    1. First derivative: dydx=c2+2c3x+3c4x2\frac{dy}{dx} = c_2 + 2c_3 x + 3c_4 x^2
    2. Second derivative: d2ydx2=2c3+6c4x\frac{d^2 y}{dx^2} = 2c_3 + 6c_4 x
    3. Third derivative: d3ydx3=6c4\frac{d^3 y}{dx^3} = 6c_4
    4. Fourth derivative: d4ydx4=0\frac{d^4 y}{dx^4} = 0 So, the required differential equation is: d4ydx4=0\frac{d^4 y}{dx^4} = 0

3. cx2+y2=4c x^2 + y^2 = 4

  • Differentiate both sides with respect to xx: 2cx+2ydydx=02cx + 2y \frac{dy}{dx} = 0 Simplify this to: cx+ydydx=0cx + y \frac{dy}{dx} = 0 This is the required differential equation.

4. y=c1ex+c2e2x+c3exy = c_1 e^{-x} + c_2 e^{2x} + c_3 e^x

  • Differentiate multiple times to eliminate constants:
    1. First derivative: dydx=c1ex+2c2e2x+c3ex\frac{dy}{dx} = -c_1 e^{-x} + 2c_2 e^{2x} + c_3 e^x
    2. Second derivative: d2ydx2=c1ex+4c2e2x+c3ex\frac{d^2 y}{dx^2} = c_1 e^{-x} + 4c_2 e^{2x} + c_3 e^x
    3. Third derivative: d3ydx3=c1ex+8c2e2x+c3ex\frac{d^3 y}{dx^3} = -c_1 e^{-x} + 8c_2 e^{2x} + c_3 e^x
    4. Fourth derivative: d4ydx4=c1ex+16c2e2x+c3ex\frac{d^4 y}{dx^4} = c_1 e^{-x} + 16c_2 e^{2x} + c_3 e^x So, the differential equation is: d4ydx43d3ydx33d2ydx2+dydx=0\frac{d^4 y}{dx^4} - 3 \frac{d^3 y}{dx^3} - 3 \frac{d^2 y}{dx^2} + \frac{dy}{dx} = 0

Would you like more details on any of these steps, or help with further clarification?


Related Questions:

  1. How can we apply the method of arbitrary constant elimination to a system of differential equations?
  2. What is the significance of higher-order derivatives in eliminating constants?
  3. How do the constants in solutions to differential equations represent initial conditions?
  4. What are the methods to solve non-homogeneous differential equations?
  5. How do boundary conditions affect the solution of a differential equation?

Tip: When eliminating arbitrary constants, always consider how many constants you have and differentiate enough times to remove all of them.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Differential Equations
Elimination of Arbitrary Constants
Calculus

Formulas

(y - 5)^2 = cx
y = c1 + c2x + c3x^2 + c4x^3
cx^2 + y^2 = 4
y = c1e^(-x) + c2e^(2x) + c3e^(x)

Theorems

Differentiation Rules
Elimination of Constants
First-Order and Higher-Order Derivatives

Suitable Grade Level

Undergraduate Calculus/Mathematics