Question

Question:(a) Use the general solution given in Example 5 to solve the IVP. 4x"+e^-0.1tx=0,x(0)=1,x'(0)=$-\frac{1}{2}$.Also use J'₀(x)=-J₁(x) and Y'₀(x)=-Y₁(x)=-Y₁(x)along withTable 6.4.1 or a CAS to evaluate coefficients.

(b) Use a CAS to graph the solution obtained in part (a) for.

Short answer

(a) The general solution is x(t) = -4.78601J₀(10e^-t/20)-3.18028Y₀(10e^-t/20) .

(b) The graph has been plotted.

Step-by-step solution

Define Bessel’s equation.

Let the Bessel equation be x²y"+xy'+(x²-n²)y=0. This equation has two linearly independent solutions for a fixed value of n . A Bessel equation of the first kind, indicated by J_n(x), is one of these solutions that may be derived usingFrobinous approach.

y₁ = x^aJ_p(bx^c)

y₁ = x^aJ_-p(bx^c)

At ,x = 0 this solution is regular. The second solution, which is singular at,x = 0 , is represented by Y_n(x) and is called a Bessel function of the second kind.

$y_3=x^a\left[\frac{\left(cosp\mathrm{\pi }\right)J_p\left(b{x}^c\right)-J_{-p}\left(b{x}^c\right)}{sinp\mathrm{\pi }}\right]$

Determine the form of the general solution.

Let the given be 4x"e^-0.1tx=0,x(0)=1,x'(0)=$\frac{-1}{2}$.That has the general solution x(t)=c₁J₀$\left(\frac{a}{2}\sqrt{\frac{k}{m}{e}^{-at/2}}\right)$+ C₂Y₀$\left(\frac{2}{a}\sqrt{\frac{k}{m}{e}^{-at/2}}\right)$, where C₁ and C₂ aare arbitrary constants

The given DE has the form of mx"+ke^-atx=0 . That yields, m= 4, k =1, a=0.1

Hence, the general solution becomes,

x(t)= C₁J0role="math" localid="1663946251828" $\left(\frac{2}{0.1}\sqrt{\frac{1}{4}{e}^{-0.1t/2}}\right)$+C₂Y₉$\left(\frac{2}{0.1}\sqrt{\frac{1}{4}{e}^{-0.1t/2}}\right)$

=C₁J₀(10e^-t/20)+C₂Y₀(10e^-t/20)

$\left(\frac{2}{a}\sqrt{\frac{k}{m}{e}^{-at/2}}\right)+c_2{Y}_0\left(\frac{2}{a}\sqrt{\frac{k}{m}{e}^{-at/2}}\right)$$\frac{-1}{2}$

Find the initial value.

Apply the initial conditions.

x(0)=C₁J₀(10e⁰)+C₂Y₀(10e⁰)

1=C₁J₀(10)+C₂Y₀(10) … (1)

x'(t)=$\frac{d}{dx}\left[C_1{J}_0\left(10e^{-t/20}\right)\right]$+$\frac{d}{dx}\left[C_2{Y}_0\left(10e^{-t/20}\right)\right]$

=

… (2)