Q20Mathematics II

Question

Q.3 Apply the method of variation of parameter to solve -

\frac{d ^{2} y}{d x ^{2}} - y = \frac{2}{1 + e ^{x}}

Answer 1

Given the second-order non-homogeneous linear ODE: $y^{''} - y = \frac{2}{1 + e ^{x}}$ .

The auxiliary equation is exactly $m^{2} - 1 = 0$ , giving distinct real roots $m = 1, - 1$ .

C.F. = C_{1} e^{x} + C_{2} e^{- x} ⟹ y_{1} = e^{x}, y_{2} = e^{- x}

We systematically calculate the fundamental Wronskian of the two independent solutions:

W = y_{1} y_{1}^{'} y_{2} y_{2}^{'} = e^{x} e^{x} e^{- x} - e^{- x} = e^{x} (- e^{- x}) - e^{x} (e^{- x}) = - 1 - 1 = - 2

The absolute Particular Integral (P.I.) is strictly assumed to be $u (x) y_{1} + v (x) y_{2}$ , where $X (x) = \frac{2}{1 + e ^{x}}$ .

u (x) = - \int \frac{y _{2} X}{W} d x = - \int \frac{e ^{- x} ( \frac{2}{1 + e ^{x}} )}{- 2} d x = \int \frac{e ^{- x}}{1 + e ^{x}} d x

To systematically integrate this, we multiply the numerator and denominator strictly by $e^{- x}$ :

u (x) = \int \frac{e ^{- 2 x}}{e ^{- x} + 1} d x

Let $t = e^{- x} + 1 ⟹ d t = - e^{- x} d x$ . Also $e^{- x} = t - 1$ .

u (x) = - \int \frac{t - 1}{t} d t = \int (\frac{1}{t} - 1) d t = ln t - t = ln (1 + e^{- x}) - (1 + e^{- x})

Since $- 1$ is simply a constant, it effectively merges strictly into the C.F., leaving $u (x) = ln (1 + e^{- x}) - e^{- x}$ .

v (x) = \int \frac{y _{1} X}{W} d x = \int \frac{e ^{x} ( \frac{2}{1 + e ^{x}} )}{- 2} d x = - \int \frac{e ^{x}}{1 + e ^{x}} d x

Let $s = 1 + e^{x} ⟹ d s = e^{x} d x$ :

v (x) = - \int \frac{1}{s} d s = - ln s = - ln (1 + e^{x})

We rigidly construct the exact P.I. = $u (x) y_{1} + v (x) y_{2}$ :

P.I. = [ln (1 + e^{- x}) - e^{- x}] e^{x} + [- ln (1 + e^{x})] e^{- x} = e^{x} ln (1 + e^{- x}) - 1 - e^{- x} ln (1 + e^{x})

y = C_{1} e^{x} + C_{2} e^{- x} - 1 + e^{x} ln (1 + e^{- x}) - e^{- x} ln (1 + e^{x})

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