Math Problem Statement

𝑦‴+𝑦″−16𝑦′+20𝑦=0;𝑦(0)=0,𝑦′(0)=1,𝑦″(0)=3.

Solution

We are given the third-order linear differential equation:

y+y16y+20y=0y''' + y'' - 16y' + 20y = 0

with the initial conditions:

y(0)=0,y(0)=1,y(0)=3y(0) = 0, \quad y'(0) = 1, \quad y''(0) = 3

Step 1: Solve the characteristic equation

The first step in solving a linear differential equation with constant coefficients is to write down the characteristic equation. For this equation, the characteristic equation is obtained by replacing yy with rr, yy' with rr, yy'' with r2r^2, and yy''' with r3r^3:

r3+r216r+20=0r^3 + r^2 - 16r + 20 = 0

Step 2: Find the roots of the characteristic equation

We will now solve the cubic equation:

r3+r216r+20=0r^3 + r^2 - 16r + 20 = 0

Let's use the Rational Root Theorem to check for possible rational roots. The possible rational roots are factors of the constant term (20) divided by factors of the leading coefficient (1), so the possible rational roots are ±1,±2,±4,±5,±10,±20\pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20.

Let's try some values. Testing r=2r = 2:

(2)3+(2)216(2)+20=8+432+20=0(2)^3 + (2)^2 - 16(2) + 20 = 8 + 4 - 32 + 20 = 0

Thus, r=2r = 2 is a root of the characteristic equation.

Step 3: Factor the cubic polynomial

Since r=2r = 2 is a root, we can factor r3+r216r+20r^3 + r^2 - 16r + 20 as:

r3+r216r+20=(r2)(r2+3r10)r^3 + r^2 - 16r + 20 = (r - 2)(r^2 + 3r - 10)

Now, solve the quadratic factor r2+3r10=0r^2 + 3r - 10 = 0 using the quadratic formula:

r=b±b24ac2ar = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} where a=1a = 1, b=3b = 3, and c=10c = -10:

r=3±(3)24(1)(10)2(1)=3±9+402=3±492r = \frac{-3 \pm \sqrt{(3)^2 - 4(1)(-10)}}{2(1)} = \frac{-3 \pm \sqrt{9 + 40}}{2} = \frac{-3 \pm \sqrt{49}}{2}

r=3±72r = \frac{-3 \pm 7}{2}

So the roots are:

r=2andr=5,2r = 2 \quad \text{and} \quad r = -5, \, 2

Step 4: General solution

The general solution to the differential equation is:

y(t)=c1e2t+c2e5t+c3e2ty(t) = c_1 e^{2t} + c_2 e^{-5t} + c_3 e^{2t}

Simplifying:

y(t)=c1e2t+c2e5t+c3e2ty(t) = c_1 e^{2t} + c_2 e^{-5t} + c_3 e^{2t}

Step 5: Apply initial conditions

Now we will use the initial conditions to solve for the constants c1c_1, c2c_2, and c3c_3.

  1. Condition y(0)=0y(0) = 0:

y(0)=c1e2(0)+c2e5(0)+c3e2(0)=0y(0) = c_1 e^{2(0)} + c_2 e^{-5(0)} + c_3 e^{2(0)} = 0 c1+c2+c3=0c_1 + c_2 + c_3 = 0

  1. Condition y(0)=1y'(0) = 1:

First, find y(t)y'(t):

y(t)=2c1e2t5c2e5t+2c3e2ty'(t) = 2c_1 e^{2t} - 5c_2 e^{-5t} + 2c_3 e^{2t} y(0)=2c15c2+2c3=1y'(0) = 2c_1 - 5c_2 + 2c_3 = 1

  1. Condition y(0)=3y''(0) = 3:

Next, find y(t)y''(t):

y(t)=4c1e2t+25c2e5t+4c3e2ty''(t) = 4c_1 e^{2t} + 25c_2 e^{-5t} + 4c_3 e^{2t} y(0)=4c1+25c2+4c3=3y''(0) = 4c_1 + 25c_2 + 4c_3 = 3

Step 6: Solve the system of equations

We now solve the system of three equations:

  1. c1+c2+c3=0c_1 + c_2 + c_3 = 0
  2. 2c15c2+2c3=12c_1 - 5c_2 + 2c_3 = 1
  3. 4c1+25c2+4c3=34c_1 + 25c_2 + 4c_3 = 3

Solving this system:

  • From the first equation, c3=c1c2c_3 = -c_1 - c_2.
  • Substitute c3c_3 into the other two equations:

2c15c2+2(c1c2)=12c_1 - 5c_2 + 2(-c_1 - c_2) = 1 2c15c22c12c2=1    7c2=1    c2=172c_1 - 5c_2 - 2c_1 - 2c_2 = 1 \implies -7c_2 = 1 \implies c_2 = -\frac{1}{7}

Now substitute c2=17c_2 = -\frac{1}{7} into c3=c1c2c_3 = -c_1 - c_2:

c3=c1+17c_3 = -c_1 + \frac{1}{7}

Substitute c2=17c_2 = -\frac{1}{7} and c3=c1+17c_3 = -c_1 + \frac{1}{7} into the third equation:

4c1+25(17)+4(c1+17)=34c_1 + 25\left(-\frac{1}{7}\right) + 4\left(-c_1 + \frac{1}{7}\right) = 3 4c12574c1+47=34c_1 - \frac{25}{7} - 4c_1 + \frac{4}{7} = 3 [ -\frac{21}{7} = 3 \implies c_3 . \

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Characteristic Equation
Cubic Equations
Initial Conditions
Linear Differential Equations with Constant Coefficients

Formulas

Characteristic Equation: r^3 + r^2 - 16r + 20 = 0
Quadratic Formula: r = (-b ± √(b^2 - 4ac)) / 2a

Theorems

Rational Root Theorem
Quadratic Formula

Suitable Grade Level

Grades 11-12