# y^{'''}-3y^{''}+2y^'=\frac{e^{2x}}{1+e^x}?

## Course: Differential Equations with Linear Algebra (don't use Linear Algebra please)

Dec 18, 2017

$y = {c}_{1} + {c}_{2} {e}^{x} + {c}_{3} {e}^{2 x} + \frac{1}{2} {e}^{x} - \frac{1}{2} L n \left({e}^{x} - 1\right) \cdot \left(2 {e}^{x} - 1\right) - \frac{1}{2} {e}^{2 x} \left(x - L n \left({e}^{x} + 1\right)\right)$

#### Explanation:

Characteristic equation of differential equation is: ${r}^{3} - 3 {r}^{2} + 2 r = 0$ or $r \cdot \left(r - 1\right) \cdot \left(r - 2\right) = 0$

Its roots are ${r}_{1} = 0$, ${r}_{2} = 1$ and ${r}_{3} = 2$

Consequently homogeneous solution of it,

${y}_{h} = {c}_{1} + {c}_{2} \cdot {e}^{x} + {c}_{3} \cdot {e}^{2 x}$

I use variation of parameters for particular solution of it,

${y}_{p} = {u}_{1} \cdot 1 + {u}_{2} \cdot {e}^{x} + {u}_{3} \cdot {e}^{2 x} = {u}_{1} + {u}_{2} \cdot {e}^{x} + {u}_{3} \cdot {e}^{2 x}$

Now, I have to solve these equation system,

$\left({u}_{1}\right) ' \cdot 1 + \left({u}_{2}\right) ' \cdot {e}^{x} + \left({u}_{3}\right) ' \cdot {e}^{2 x} = 0$ or $\left({u}_{1}\right) ' + \left({u}_{2}\right) ' \cdot {e}^{x} + \left({u}_{3}\right) ' \cdot {e}^{2 x} = 0$ $\left(1\right)$

$\left({u}_{1}\right) ' \cdot 0 + \left({u}_{2}\right) ' \cdot {e}^{x} + \left({u}_{3}\right) ' \cdot 2 {e}^{2 x} = 0$ or $\left({u}_{2}\right) ' \cdot {e}^{x} + 2 \left({u}_{3}\right) ' \cdot {e}^{2 x} = 0$ $\left(2\right)$

$\left({u}_{1}\right) ' \cdot 0 + \left({u}_{2}\right) ' \cdot {e}^{x} + \left({u}_{3}\right) ' \cdot 4 {e}^{2 x} = {e}^{2 x} / \left({e}^{x} + 1\right)$ or $\left({u}_{2}\right) ' \cdot {e}^{x} + 4 \left({u}_{3}\right) ' \cdot {e}^{2 x} = {e}^{2 x} / \left({e}^{x} + 1\right)$ $\left(3\right)$

$\left[\left({u}_{2}\right) ' \cdot {e}^{x} + 4 \left({u}_{3}\right) ' \cdot {e}^{2 x}\right] - \left[\left({u}_{2}\right) ' \cdot {e}^{x} + 2 \left({u}_{3}\right) ' \cdot {e}^{2 x}\right] = {e}^{2 x} / \left({e}^{x} + 1\right) - 0$

$2 \left({u}_{3}\right) ' \cdot {e}^{2 x} = {e}^{2 x} / \left({e}^{x} + 1\right)$

$\left({u}_{3}\right) ' = \frac{1}{2} \cdot \frac{1}{{e}^{x} + 1}$

${u}_{3} = \frac{1}{2} \int \frac{\mathrm{dx}}{{e}^{x} + 1}$

=$\frac{1}{2} \int \frac{{e}^{- x} \cdot \mathrm{dx}}{{e}^{- x} + 1}$

=$\frac{1}{2} L n \left({e}^{- x} + 1\right)$

=$\frac{1}{2} L n \left(\frac{{e}^{x} + 1}{e} ^ x\right)$

=$\frac{1}{2} L n \left({e}^{x} + 1\right) - \frac{1}{2} L n \left({e}^{x}\right)$

=$\frac{1}{2} L n \left({e}^{x} + 1\right) - \frac{x}{2}$

Consequently,

$\left({u}_{2}\right) ' \cdot {e}^{x} + 2 \left({u}_{3}\right) ' \cdot {e}^{2 x} = 0$

$\left({u}_{2}\right) ' = - 2 {e}^{x} \cdot \left({u}_{3}\right) '$

$\left({u}_{2}\right) ' = - 2 {e}^{x} \cdot \frac{1}{2} \cdot \frac{1}{{e}^{x} + 1}$

$\left({u}_{2}\right) ' = - {e}^{x} / \left({e}^{x} + 1\right)$

${u}_{2} = - L n \left({e}^{x} + 1\right)$

Hence,

$\left({u}_{1}\right) ' - {e}^{x} / \left({e}^{x} + 1\right) \cdot {e}^{x} + \frac{1}{2} \cdot \frac{1}{{e}^{x} + 1} \cdot {e}^{2 x} = 0$

$\left({u}_{1}\right) ' - {e}^{2 x} / \left({e}^{x} + 1\right) + \frac{1}{2} \cdot {e}^{2 x} / \left({e}^{x} + 1\right) = 0$

$\left({u}_{1}\right) ' = \frac{1}{2} \cdot {e}^{2 x} / \left({e}^{x} + 1\right)$

${u}_{1} = \frac{1}{2} \int \frac{{e}^{2 x} \cdot \mathrm{dx}}{{e}^{x} + 1}$

=$\frac{1}{2} \int \frac{\left({e}^{x} + 1\right) \cdot \left({e}^{x} - 1\right) + 1}{{e}^{x} + 1} \cdot \mathrm{dx}$

=$\frac{1}{2} \int \left({e}^{x} - 1\right) \cdot \mathrm{dx}$+$\frac{1}{2} \int \frac{\mathrm{dx}}{{e}^{x} + 1}$

=$\frac{1}{2} \cdot \left({e}^{x} - x\right) + \frac{1}{2} \cdot \left(L n \left({e}^{x} + 1\right) - x\right)$

=$\frac{1}{2} {e}^{x} - \frac{x}{2} + \frac{1}{2} L n \left({e}^{x} + 1\right) - \frac{x}{2}$

=$\frac{1}{2} {e}^{x} + \frac{1}{2} L n \left({e}^{x} + 1\right) - x$

Hence,

${y}_{p} = {u}_{1} + {u}_{2} \cdot {e}^{x} + {u}_{3} \cdot {e}^{2 x}$

=$\frac{1}{2} {e}^{x} + \frac{1}{2} L n \left({e}^{x} + 1\right) - x - {e}^{x} L n \left({e}^{x} + 1\right) + \frac{1}{2} {e}^{2 x} \cdot L n \left({e}^{x} + 1\right) - \frac{x}{2} \cdot {e}^{2 x}$

=$\frac{1}{2} {e}^{x} - \frac{1}{2} L n \left({e}^{x} - 1\right) \cdot \left(2 {e}^{x} - 1\right) - \frac{1}{2} {e}^{2 x} \left(x - L n \left({e}^{x} + 1\right)\right)$

Thus,

$y = {y}_{h} + {y}_{p}$

=${c}_{1} + {c}_{2} {e}^{x} + {c}_{3} {e}^{2 x} + \frac{1}{2} {e}^{x} - \frac{1}{2} L n \left({e}^{x} - 1\right) \cdot \left(2 {e}^{x} - 1\right) - \frac{1}{2} {e}^{2 x} \left(x - L n \left({e}^{x} + 1\right)\right)$