Toronto Math Forum

Find a particular solution of the given inhomogeneous Euler's euqation:
    (1 - t)$y\prime\prime$ + t$y\prime$ - y = $2(t - 1)^2$$e^{-t}$, 0 < t < 1, $ y_1(t) $=$ e^t$.

Solution:

$y_1$ = $e^{t}$,  $y_1\prime$ = $e^{t}$

Rewrite:

$y\prime\prime$ + $\frac{t}{1-t}$$y\prime$ - $\frac{y}{1 - t}$ = $\frac{2(t-1)^2e^{-t}}{1-t}$

p(t) = $\frac{t}{1-t}$

g(t) = 2$\frac{(t-1)^2e^{-t}}{1-t}$

$y_1$$ v\prime\prime$ + [2$y_1\prime$ + p(t)y_1]$v\prime$ = g(t)

$e^t$$v\prime\prime$ + [2$e^t$+ $\frac{t}{1-t}$$e^t$]$v\prime$ = 2$\frac{(t-1)^2e^{-t}}{1-t}$

$v\prime\prime$ + [$\frac{2-t}{1-t}$]$v\prime$ = 2$\frac{(t-1)^2e^{-t}}{1-t}$

Let w = $v\prime$:

$w\prime$ + [$\frac{2-t}{1-t}$]w = 2$\frac{(t-1)^2e^{-t}}{1-t}$

$\mu(t)$ = $e^{\int \frac{2-t}{1-t}}$ = $\frac{e^t}{1-t}$

$\frac{e^t}{1-t}$$w\prime$ + $\frac{e^t}{1-t}$[$\frac{2-t}{1-t}$]w = 2$\frac{(t-1)^2e^{-t}}{1-t}$$\frac{e^t}{1-t}$

$\mu(t)$w = $\int \frac{2(t-1)^2e^{-t}}{(1-t)^2}$ = $\int 2e^{-t}$ = $-2e^{-t}$

w = -2$e^{-t}$$\frac{1-t}{e^t}$ = -2(1-t)$e^{-2t}$

w = $v\prime$ = -2(1-t)$e^{-2t}$

v = $\int -2(1-t)e^{-2t}$

By using substitution, we get:
v = (1-t)$e^{-2t}$- $\frac{1}{2}$$ e^{-2t}$

Y(t) = v$y_1$ = (1-t) $e^{-t}$- $\frac{1}{2}$$e^{-2t}$