Problem 28 Find for $x=1$ the $y$ value... [FREE SOLUTION]

Chapter 6: Problem 28

Find for $x=1$ the $y$ value for the particular solution required. $$ \left(D^{2}+3 D-10\right) y=0 ; \text { when } x=0, y=0, \text { and when } x=2, y=1. $$

Short Answer

Expert verified

The value of $y$ at $x=1$ is $\frac{e^4 - e^{-7}}{e^8 - e^{-14}}$.\)

Step by step solution

Identify the characteristic equation

The differential equation given is $(D^2 + 3D - 10)y = 0$. The corresponding characteristic equation is $r^2 + 3r - 10 = 0$.

Solve the characteristic equation

Solve $r^2 + 3r - 10 = 0$ using the quadratic formula: $r = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$, where $a = 1$, $b = 3$, and $c = -10$.

Calculate the roots

Substitute values into the quadratic formula: \[r = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-10)}}{2 \cdot 1}\]\[r = \frac{-3 \pm \sqrt{9 + 40}}{2}\]\[r = \frac{-3 \pm \sqrt{49}}{2}\]We find the roots are $r_1 = 4$ and $r_2 = -7$.

Write the general solution

Since the roots are distinct, the general solution is $y(x) = C_1 e^{4x} + C_2 e^{-7x}$.

Apply the initial conditions

Use the initial conditions $y(0) = 0$ and $y(2) = 1$ to find the values of the constants $C_1$ and $C_2$.

Substitute the first condition

Substitute $x = 0$ into the general solution:$0 = C_1 e^0 + C_2 e^0$ $0 = C_1 + C_2$, giving $C_2 = -C_1$.

Substitute the second condition

Substitute $x = 2$ into the equation and simplify:$1 = C_1 e^{4 \cdot 2} + C_2 e^{-7 \cdot 2}$$1 = C_1 e^8 - C_1 e^{-14}$,so $C_1(e^8 - e^{-14}) = 1$.

Solve for C1

Rearrange to find $C_1$:\[C_1 = \frac{1}{e^8 - e^{-14}}\].

Solve for C2

Substitute $C_1$ into $C_2=-C_1$: \[C_2 = -\frac{1}{e^8 - e^{-14}}\].

Find the particular solution at x=1

Substitute $x = 1$ into $y(x)=C_1 e^{4x}+C_2 e^{-7x}$:\[y(1)=\frac{1}{e^8 - e^{-14}}(e^4 - e^{-7})\].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Characteristic Equation

In differential equations, the characteristic equation is a crucial concept for solving linear differential equations with constant coefficients. It is essentially a polynomial equation derived directly from the differential operator. Here's how you form it:

Start with the differential equation. For example, $(D^2 + 3D - 10)y = 0$.
Replace each derivative operator $D^n$ with $r^n$; thus, it converts the differential expression into a polynomial.
This gives the characteristic equation $r^2 + 3r - 10 = 0$.

Once you've formed the characteristic equation, solving it gives you the roots $r$, which are central to forming the general solution of the differential equation.

Quadratic Formula

The quadratic formula is a standard method for finding the roots of a quadratic equation $ax^2 + bx + c = 0$. You might remember this formula from algebra classes:\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

Here, $a$, $b$, and $c$ are the coefficients from the quadratic equation.
For our characteristic equation $r^2 + 3r - 10 = 0$, $a = 1$, $b = 3$, and $c = -10$.

By substituting these values into the quadratic formula, we find the roots. This step is crucial because determining the roots will guide us in constructing the general solution of the differential equation.

Initial Conditions

Initial conditions are specific values provided in a differential equation problem to determine the constants in its general solution. They help us find a particular solution, which satisfies the differential equation and aligns with real-world situations, like population growth or temperature changes over time.

In this exercise, the initial conditions given are $y(0) = 0$ and $y(2) = 1$.
Initial conditions allow us to replace arbitrary constants $C_1$ and $C_2$ with specific values in the general solution.

Apply these conditions to the general solution to solve for $C_1$ and $C_2$. Each condition lets you set up an equation, which you then solve to find these constants.

General Solution

The general solution of a differential equation contains all possible solutions and includes arbitrary constants. For a second-order differential equation, it usually involves two arbitrary constants.Based on the distinct roots $r_1 = 4$ and $r_2 = -7$ from the characteristic equation, our general solution forms as:\[y(x) = C_1 e^{4x} + C_2 e^{-7x}\]

This form results specifically from combining the solutions corresponding to each root.
The constants $C_1$ and $C_2$ hybridize the exponential functions, tailoring the general solution to meet the initial conditions specific to the problem.

Once you apply the initial conditions, you will solve for these constants, turning the general solution into a particular solution that fits the scenario described by the exercise.

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91影视

Find for \(x=1\) the \(y\) value for the particular solution required. $$ \left(D^{2}+3 D-10\right) y=0 ; \text { when } x=0, y=0, \text { and when } x=2, y=1. $$