Question

Solve the differential equation in Problem 27 if the boundary conditions are T(1)=160, T(3)=90.

Short answer

$T\left(r\right)=70+\frac{\left(90K_0\right(3a)-20K_0(a\left)\right)I_0\left(ar\right)+\left(20I_0\left(a\right)-90I_0\left(3a\right)\right)K_0\left(ar\right)}{I_0\left(a\right)K_0\left(3a\right)-K_0\left(a\right)I_0\left(3a\right)}$

Step-by-step solution

Definition of Power series solutions

If x=x₀is an ordinary point of the differential equation , we can always find two linearly independent solutions in the form of a power series centered at x₀, that is,

$y=\sum _{n=0}^x{C}_n{(X-X_0)}^n$

A power series solution converges at least on some interval defined by$\left|X-X_0\right|<R$, where R is the distance from x₀ to the closest singular point.

Using Boundary conditions to solve the problem

As in the previous exercise we introduce the substitution

w(r)=T(r)-70.

The new boundary conditions are

w(1)=T(1)-70=160-70=90,w(3)=T(3)-70=90-70=20.

Since the original equation is the same we know that the general solution is given by

w(r)c₁I₀(ar)+c₂K₀(ar)

The initial conditions lead us to the following system of equations

c₁I₀(a)+c₂K₀(a)=90

c₁I₀(3a)+c₂K₀(3a)=20

Using Kramer’s rule

If we solve this system via Kramer's rule we see that the determinant of the system is

$D=\left|\begin{array}{cc}{I}_0\left(a\right)& K_0\left(a\right)\\ I_0\left(3a\right)& K_0\left(3a\right)\end{array}\right|=I_0\left(a\right)K_0\left(3a\right)-K_0\left(a\right)I_0\left(3a\right)$

And

$$\begin{gathered} D_1=\left|\begin{array}{cc}90& K_0\left(a\right)\\ 20& K_0\left(3a\right)\end{array}\right|=90K_0\left(3a\right)-20K_0\left(a\right) \\ {D}_2=\left|\begin{array}{cc}{l}_0\left(a\right)& 90\\ I_0\left(3a\right)& 20\end{array}\right|=20I_0\left(a\right)-90I_0\left(3a\right) \end{gathered}$$

We get that

$C_1=\frac{D_1}{D}=\frac{90K_0\left(3a\right)-20K_0\left(a\right)}{I_0\left(a\right)K_0\left(3a\right)-K_0\left(a\right)I_0\left(3a\right)}$

and

$C_2=\frac{D_1}{D}=\frac{20I_0\left(a\right)-90I_0\left(3a\right)}{I_0\left(a\right)K_0\left(3a\right)-K_0\left(a\right)I_0\left(3a\right)}$

We conclude that the solution of the problem foris given by

role="math" localid="1663919005221" $w\left(r\right)=\frac{\left(90K_0\right(3a)-20K_0(a\left)\right)I_o\left(ar\right)+\left(20I_0\right(a)-90I_0(3a\left)\right)K_0\left(ar\right)}{I_0\left(a\right)K_0\left(3a\right)-K_0\left(a\right)I_0\left(3a\right)}$

Reversing the substitution we get that the solution of the initial problem is

$T\left(r\right)=70+\frac{\left(90K_0\right(3a)-20K_0(a\left)\right)I_o\left(ar\right)+\left(20I_0\right(a)-90I_0(3a\left)\right)K_0\left(ar\right)}{I_0\left(a\right)K_0\left(3a\right)-K_0\left(a\right)I_0\left(3a\right)}$