Answer to Question #140451 in Differential Equations for Anurag Dhayal

Question #140451
y"+2y'+y=3e^-x√x+1 find the general solution
1
Expert's answer
2020-10-27T18:38:06-0400

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

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

k2+2k+1=0(k+1)2=0k1=k2=1\begin{array}{l} {k^2} + 2k + 1 = 0\\ {(k + 1)^2} = 0\\ {k_1} = {k_2} = - 1 \end{array}

Therefore yc=C1ex+xC2ex{y_c} = {C_1}{e^{ - x}} + x{C_2}{e^{ - x}}

Let

C1=f(x),C2=g(x){C_1} = f(x),\,{C_2} = g(x)

Then

y=fex+xgexy=fexfex+g(exxex)+gxexy = f{e^{ - x}} + xg{e^{ - x}} \Rightarrow y' = f'{e^{ - x}} - f{e^{ - x}} + g({e^{ - x}} - x{e^{ - x}}) + g'x{e^{ - x}}

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

Then

y=fex+g(exxex)y=fex+fex+g(exxex)+g(2ex+xex)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:

fex+fex+g(exxex)+g(2ex+xex)+2(fex+g(exxex))++fex+xgex=3exx+1\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} fex+g(exxex)=3exx+1- f'{e^{ - x}} + g'({e^{ - x}} - x{e^{ - x}}) = 3{e^{ - x}}\sqrt {x + 1}

with(*) we have

{fex+g(exxex)=3exx+1fex+gxex=0\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=gxf' = - g'x

Then

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

f=gx=3xx+1f=25(3x2)(x+1)x+1+C2y=fex+xgex=(25(3x2)(x+1)x+1+C2)ex+x(2x+1(x+1)+C)ex\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}



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