Consider the ODE
y'(x) = -\dfrac{y(x)}{x - A} + \dfrac{Ex}{x - A}
with y(0) = D and x > A.
Determine the value \color{orange} y(C)
of the solution function of the IVP.
\color{orange} y(C)
=
(0.5*E * C*C - A*D)/(C-A)
First, we determine the general solution of the associated homogeneous ODE y'(x) = p(x) \cdot y(x).
It is given by y(x) = K \cdot e^{P(x)}.
Here, K is a constant and \displaystyle P(x) \in \int p(x) dx is an antiderivative of p.
We have \displaystyle P(x) \in \int p(x) dx =
\int -\dfrac{1}{x - A} dx =
-\ln |x - A| = -\ln(x - A) + constant, since x > A.
Substituting gives
y(x) = K \cdot e^{P(x)} = K \cdot e^{-\ln(x - A)} = K \cdot \dfrac{1}{x - A}.
The integration constant
K is determined in the end by the initial condition y(0) = D, along with other constants.
Using the method of variation of constants, the general solution of the inhomogeneous ODE is
\displaystyle y(x) = \left( \int e^{-P(x)} \cdot q(x) dx + C
\right) e^{P(x)}.
We calculate e^{-P(x)} \cdot q(x) = e^{-(-\ln(x - A))} \cdot \dfrac{Ex}{x - A} =
(x - A) \cdot \dfrac{Ex}{x - A} = Ex
and \displaystyle \int e^{-P(x)} \cdot q(x) dx = \int Ex dx = fractionReduce(E,2) x^2 + C.
The general solution of the inhomogeneous ODE is therefore
\displaystyle y(x) = \left(fractionReduce(E,2) x^2 + C
\right) e^{P(x)} = \left(fractionReduce(E,2) x^2 + C
\right) \frac{1}{x - A}
= \frac{fractionReduce(E,2) x^2 + C}{x - A}.
With
y(0) = D = \dfrac{0 + C}{0 - A}, we find that C = -A*D.
Thus, the solution of the IVP is \displaystyle y(x) = \frac{fractionReduce(E,2) x^2 + -A*D}{x - A}.
Substitution gives {\color{orange} y(C)} =
fractionReduce(0.5*E * C*C - A*D, C-A)
.