Question

y"+2y'+y=3e^-x√x+1 find the general solution

Answer 1

$y'' + 2y' + y = 3{e^{ - x}}\sqrt {x + 1}$

Let's find ${y_c}$ , using auxiliary equation:

$\begin{array}{l} {k^2} + 2k + 1 = 0\\ {(k + 1)^2} = 0\\ {k_1} = {k_2} = - 1 \end{array}$

Therefore ${y_c} = {C_1}{e^{ - x}} + x{C_2}{e^{ - x}}$

Let

${C_1} = f(x),\,{C_2} = g(x)$

Then

$y = f{e^{ - x}} + xg{e^{ - x}} \Rightarrow y' = f'{e^{ - x}} - f{e^{ - x}} + g({e^{ - x}} - x{e^{ - x}}) + g'x{e^{ - x}}$

$\begin{array}{l} Let\\ f'{e^{ - x}} + g'x{e^{ - x}} = 0\,\,\,(*) \end{array}$

Then

$y' = - f{e^{ - x}} + g({e^{ - x}} - x{e^{ - x}}) \Rightarrow y'' = - f'{e^{ - x}} + f{e^{ - x}} + g'({e^{ - x}} - x{e^{ - x}}) + g( - 2{e^{ - x}} + x{e^{ - x}})$

Substituting into our equation for y, y', y'', we have:

$\begin{array}{l} - f'{e^{ - x}} + f{e^{ - x}} + g'({e^{ - x}} - x{e^{ - x}}) + g( - 2{e^{ - x}} + x{e^{ - x}}) + 2\left( { - f{e^{ - x}} + g({e^{ - x}} - x{e^{ - x}})} \right) + \\ + f{e^{ - x}} + xg{e^{ - x}} = 3{e^{ - x}}\sqrt {x + 1} \end{array}$ $- f'{e^{ - x}} + g'({e^{ - x}} - x{e^{ - x}}) = 3{e^{ - x}}\sqrt {x + 1}$

with(*) we have

$\left\{ \begin{array}{l} - f'{e^{ - x}} + g'({e^{ - x}} - x{e^{ - x}}) = 3{e^{ - x}}\sqrt {x + 1} \\ f'{e^{ - x}} + g'x{e^{ - x}} = 0\, \end{array} \right.$

From 2-nd equation:

$f' = - g'x$

Then

$g'x + g' - g'x = 3\sqrt {x + 1} \Rightarrow g' = 3\sqrt {x + 1} \Rightarrow g = 2\sqrt {x + 1} (x + 1) + C$

$\begin{array}{l} f' = - g'x = - 3x\sqrt {x + 1} \Rightarrow f = - \frac{2}{5}\left( {3x - 2} \right)\left( {x + 1} \right)\sqrt {x + 1} + {C_2}\\ y = f{e^{ - x}} + xg{e^{ - x}} = \left( { - \frac{2}{5}\left( {3x - 2} \right)\left( {x + 1} \right)\sqrt {x + 1} + {C_2}} \right){e^{ - x}} + x\left( {2\sqrt {x + 1} (x + 1) + C} \right){e^{ - x}} \end{array}$