Answer to Question #288024 in Differential Equations for Abhishek

Question

(D^2-3D+2)y= e^x/1+e^x

Answer 1

Characteristic equation

${k^2} - 3k + 2 = 0$

$(k - 1)(k - 2) = 0$

${k_1} = 1;\,{k_2} = 2$

We obtain the general solution of the homogeneous equation

${y_0} = {C_1}{e^x} + {C_2}{e^{2x}}$

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

Then

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

Let

$f'{e^x} + g'{e^{2x}} = 0\,\,\,(*)$

Then

$y' = f{e^x} + 2g{e^{2x}} \Rightarrow y'' = f'{e^x} + f{e^x} + 2g'{e^{2x}} + 4g{e^{2x}}$

Substitute in the original equation

$f'{e^x} + f{e^x} + 2g'{e^{2x}} + 4g{e^{2x}} - 3\left( {f{e^x} + 2g{e^{2x}}} \right) + 2\left( {f{e^x} + g{e^{2x}}} \right) = \frac{{{e^x}}}{{1 + {e^x}}}$

$f'{e^x} + 2g'{e^{2x}} = \frac{{{e^x}}}{{1 + {e^x}}}$

Together with (*) we have a system of equations

$\left\{ \begin{array}{l} f'{e^x} + 2g'{e^{2x}} = \frac{{{e^x}}}{{1 + {e^x}}}\\ f'{e^x} + g'{e^{2x}} = 0 \end{array} \right.$

from 2-nd eauation:

$f' = - g'{e^x}$

Substitute in the 1-st equation

$- g'{e^{2x}} + 2g'{e^{2x}} = \frac{{{e^x}}}{{1 + {e^x}}}$

$g' = \frac{{{e^{ - x}}}}{{1 + {e^x}}}$

$g = \int {\frac{{{e^{ - x}}}}{{1 + {e^x}}}dx = - x - {e^{ - x}}} + \ln ({e^x} + 1) + {C_1}$

$f' = - g'{e^x} = - \frac{1}{{1 + {e^x}}} \Rightarrow f = - \int {\frac{1}{{1 + {e^x}}}} dx = \ln ({e^x} + 1) - x + {C_2}$

Then

$y = f{e^x} + g{e^{2x}} = \left( {\ln ({e^x} + 1) - x + {C_2}} \right){e^x} + \left( { - x - {e^{ - x}} + \ln ({e^x} + 1) + {C_1}} \right){e^{2x}}$

Answer: $y = \left( {\ln ({e^x} + 1) - x + {C_2}} \right){e^x} + \left( { - x - {e^{ - x}} + \ln ({e^x} + 1) + {C_1}} \right){e^{2x}}$