Math Problem Statement

Solve the given initial value problem.

y double prime plus 5 y primey′′+5y′equals=​0;

         

​y(0)equals=negative 4−4​,

    

y prime left parenthesis 0 right parenthesisy′(0)equals=negative 5−5

Question content area bottom

Part 1

The solution is

​y(t)equals=enter your response here.

Solution

We are given the second-order linear differential equation:

y+5y=0y'' + 5y' = 0 with initial conditions:

y(0)=4,y(0)=5.y(0) = -4, \quad y'(0) = -5.

Step 1: Solve the homogeneous equation

The given differential equation is:

y+5y=0.y'' + 5y' = 0.

This is a linear homogeneous differential equation with constant coefficients. The characteristic equation associated with this differential equation is:

r2+5r=0.r^2 + 5r = 0.

We can factor this equation:

r(r+5)=0.r(r + 5) = 0.

So, the roots are:

r=0,r=5.r = 0, \quad r = -5.

Thus, the general solution to the differential equation is:

y(t)=C1e0t+C2e5t.y(t) = C_1 e^{0t} + C_2 e^{-5t}.

Since e0t=1e^{0t} = 1, we simplify this to:

y(t)=C1+C2e5t.y(t) = C_1 + C_2 e^{-5t}.

Step 2: Use the initial conditions to determine C1C_1 and C2C_2

  1. Apply the first initial condition y(0)=4y(0) = -4:

Substitute t=0t = 0 into the solution:

y(0)=C1+C2e0=C1+C2=4.y(0) = C_1 + C_2 e^{0} = C_1 + C_2 = -4.

Thus, we get the equation:

C_1 + C_2 = -4. \tag{1}

  1. Apply the second initial condition y(0)=5y'(0) = -5:

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

y(t)=ddt(C1+C2e5t)=5C2e5t.y'(t) = \frac{d}{dt} \left( C_1 + C_2 e^{-5t} \right) = -5C_2 e^{-5t}.

Substitute t=0t = 0:

y(0)=5C2e0=5C2=5.y'(0) = -5C_2 e^{0} = -5C_2 = -5.

So, we get:

C_2 = 1. \tag{2}

Step 3: Solve for C1C_1

Substitute C2=1C_2 = 1 into equation (1):

C1+1=4,C_1 + 1 = -4, which gives:

C1=5.C_1 = -5.

Step 4: Write the final solution

Now that we have C1=5C_1 = -5 and C2=1C_2 = 1, the solution to the differential equation is:

y(t)=5+e5t.y(t) = -5 + e^{-5t}.

Final Answer:

y(t)=5+e5t.y(t) = -5 + e^{-5t}.

Would you like further explanation on any part of the solution? Or do you have any other questions?

Related Questions:

  1. What is the behavior of the solution as tt \to \infty?
  2. How would the solution change if the equation had a different constant in front of yy'?
  3. Can this method be applied to higher-order linear differential equations?
  4. How do you solve non-homogeneous differential equations with similar methods?
  5. What is the significance of the initial conditions in determining the constants C1C_1 and C2C_2?

Tip: Always check if the roots of the characteristic equation are real or complex, as this determines the form of the solution (exponential vs. sinusoidal).

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Differential Equations
Initial Value Problems
Homogeneous Linear Differential Equations
Characteristic Equations

Formulas

Characteristic Equation: r^2 + 5r = 0
General Solution: y(t) = C1 e^(r1 t) + C2 e^(r2 t)
Derivative Rule: d/dt (C e^(at)) = a C e^(at)

Theorems

Theory of Linear Differential Equations with Constant Coefficients
Method of Solving Homogeneous Differential Equations

Suitable Grade Level

Undergraduate Calculus (Differential Equations)