}

# Preface

The Laplace transform helps also to solve variable coefficient linear differential equations.

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# Nonconstant coefficient IVP’s

Example: Consider the initial value problem

$\left( t +1\right) \ddot{y} - \dot{y} -t\,y(t) = 0, \qquad y(0) =1, \quad \dot{y} = -1.$
where overdot stands for the derivative with respect to t.    ■

Example: Consider the initial value problem

$t\,\ddot{y} - \dot{y} -t\,y(t) = 0, \qquad y(0) =1, \tag{Example2.1}$
where overdot stands for the derivative with respect to t. Note that this IVP has only one initial condition because the differential equation has a regular singular point at t = 0. The given differential equation has two linearly independent solution, but only one of them is of exponential order and admits Laplace transformation. Another linearly independent solution is underfined at the origin and is not suitable for our technique. So we have to disregard this solution and work only with one of them. Therefore, only one initial condition is relevant in our case.

First, we ask Mathematica whether it knows its general solution.

DSolve[t*y''[t] - y'[t] - t*y[t] == 0, y[t], t]
{{y[t] -> -t BesselJ[1, I t] C[1] + t BesselY[1, -I t] C[2]}}
It turns out that the general solution is expressed through modified Bessel functions I1 and K1:
$y(t) = C_1 t\,I_1 (t) + C_2 t\,K_1 (t) ,$
where C1 and C1 are arbitrary constants. Next, we find the limits when the argument tends to zero (in order to verify the initial condition):
Limit[x*BesselI[1, x], x -> 0]
0
Limit[x*BesselK[1, x], x -> 0]
1
At infinity, the modified Bessel function of the first kind is growing exponentially:
$t\,I_1 (t) \sim \sqrt{\frac{t}{2\pi}} \,e^t \qquad\mbox{as} \quad t \to \infty .$
So it is of exponential order and its Laplace transform exists and equals
${\cal L}_{t\to\lambda} \left[ t\,I_1 (t) \right] = \left( \lambda^2 -1 \right)^{-3/2} .$
Another function involving the Kelvin function is bounded ($$\left\vert t\,K_1 (t) \right\vert \le 1$$ for 0 ≤ t <∞) and its Laplace transform exists, but the corresponding explicit formula is unknown yet.

The initial condition is
$y(0) = C_1 \cdot 0 + C_2 \cdot 1 = C_2 ,$
but the first derivative
$\dot{y}(0) = C_2 \lim_{t\to 0} \left[ K_1 (t) - t\,K_2 (t) \right] \qquad \mbox{undetermined}.$
Therefore, we cannot specify the second initial condition for the given singular differential equation. Now we can make a conclusion that the given inital value problem has a solution only when the initial condition is homogeneous:
$t\,\ddot{y} - \dot{y} -t\,y(t) = 0, \qquad y(0) =0. \tag{Example2.2}$
Its solution is not unique:
$y(t) = C_1 t\, I_1 (t) , \tag{Example2.3}$
and depends on an arbitrary constant C1. On the other hand, the formulkated initially the problem (Example2.1) has no solution. Now we verify this conclusion with the aid of the Laplace transform.

We need to determine the Laplace transforms of each term in the given differential equation. So we have

\begin{align*} {\cal L}\left[ y(t) \right] &= y^L (\lambda ) = \int_0^{\infty} y(t) \,e^{-\lambda t} {\text d} t , \\ {\cal L}\left[ t\,y(t) \right] &= \int_0^{\infty} t\,y(t) \,e^{-\lambda t} {\text d} t = - \frac{\text d}{{\text d}\lambda} \, y^L (\lambda ) , \\ {\cal L}\left[ \dot{y}(t) \right] &= \lambda\,y^L - y(0) = \lambda\,y^L -1. \end{align*}
Integrating by parts, we get
\begin{align*} {\cal L}\left[ t\ddot{y}(t) \right] &= \int_0^{\infty} t\ddot{y}(t) \, e^{-\lambda t} {\text d} t = - \int_0^{\infty} \dot{y} \left[ e^{-\lambda t} - \lambda t\,e^{-\lambda t} \right] {\text d} t \\ &= y(0) \left[ e^{-\lambda t} - \lambda t\,e^{-\lambda t} \right]_{t=0} + \int_0^{\infty} y(t) \left[ -2\lambda\, e^{-\lambda t} + \lambda^2 t \,e^{-\lambda t} \right] {\text d} t \\ &= y(0) -2\lambda\,y^L + \lambda^2 {\cal L} \left[ t\,y(t) \right] . \end{align*}
Putting all terms together, we obtain the equation for the Laplace transform of the required function:
$1 - 2\lambda\,y^L -\lambda^2 \frac{\text d}{{\text d}\lambda} \, y^L (\lambda ) -\lambda\,y^L + 1 + \frac{\text d}{{\text d}\lambda} \, y^L (\lambda ) =0,$
or upon simplification,
$\left( 1 - \lambda^2 \right) \frac{\text d}{{\text d}\lambda} \, y^L (\lambda ) - 3\lambda\,y^L = -2 .$
Its solution is
DSolve[(x^2 - 1)*y'[x] + 3*y[x] == 2, y, x]
{{y -> Function[{x}, ( 2 E^(-3 (1/2 Log[1 - x] - 1/2 Log[1 + x])) (1 - x)^(3/2))/( 3 (1 + x)^(3/2)) + E^(-3 (1/2 Log[1 - x] - 1/2 Log[1 + x])) C[1]]}}
$y^L (\lambda ) = \frac{c}{\left( \lambda^2 -1 \right)^{3/2}} + \frac{2}{3} \left( 1 - \lambda \right)^{-3/2} ,$
where c is a constant of integration. Taking the inverse Laplac e transform, we obtain
$y(t) = ct\,I_1 (t) + \frac{1}{3\sqrt{2}}\,\delta (t) ,$
where I1(t) is the modofied Bessel function of order 1 and δ is the delta Dirac function.
InverseLaplaceTransform[(p - 1)^(-3/2) , p, t]
DiracDelta[t]/(2 Sqrt[2])
■

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