Answer to Question #127129 in Differential Equations for Damba

Question

Find the general solution of
y" + 3y' + 2y = 1/(1+e^x)

Answer 1

We solve a second order differential equation of the type:

ay''+by'+cy=f(x)

where a=1

b=3

c=2

f(x)=1/(1+e^x)

The solution will consist of common and particular solutions and have a type:

y_gen=y_com+y_part

To find a common solution we solve second order differential equation y''+3y'+2y=0 and find roots of polynomial r²+3r+2=0

Its roots  

$r_1={{-3+\sqrt{3^2-4\sdot1\sdot2}}\over{2\sdot1}}={{{-3+\sqrt{1}}\over{2}}={{-3+1}\over{2}}}={-1}$

$r_2={{-3-\sqrt{3^2-4\sdot1\sdot2}}\over{2\sdot1}}={{{-3-\sqrt{1}}\over{2}}={{-3-1}\over{2}}}={-2}$

​Roots are complex therefore the common solution will be $y_{com}=C_1e^{-{x}}+C_2e^{-2{x}}$

We will use the method of variation of parameters to find a particular solution

$\begin{cases} C'_1(x)y_1+C'_2(x)y_2=0 \\ C'_1(x)y'_1+C'_2(x)y'_2={f(x)\over {a_0}} \end{cases}$

$y_1=e^{-{x}}\\ y_2=e^{-{2x}}\\ a_0=1$

finally we get

$\begin{cases} C'_1(x)e^{-{x}}+C'_2(x)e^{-{2x}}=0 (1)\\ -C'_1(x)e^{-{x}}-2C'_2(x)e^{-{2x}}={1\over {1+e^{{x}}}} (2) \end{cases}$

add (2) to (1) and we get

$-2C'_2(x)e^{-{2x}}={1\over {1+e^{x}}}\\ C'_2(x)=-{e^{2x}\over {1+e^{x}}}$

We will make some changes and simplifications

$C'_2(x)=-{e^{2x}+e^{x}-e^{x}\over {1+e^{x}}}=-{e^{x}(e^{x}+1)\over {e^{x}+1}}+{e^{x}\over {e^{x}+1}}=-e^{x}+{e^{x}\over {e^{x}+1}}$

Let us integrate C'₂

$C_2(x)=\lmoustache{(-e^{x}+{e^{x}\over {e^{x}+1}})dx}=-e^{x}+ln\vert{{e^{x}+1}}\vert+C_2$

we multiply (1) by 2, add to (2) and get

$C'_1(x)e^{-{x}}={1\over {1+e^{x}}}\\ C'_1(x)={e^{x}\over {1+e^{x}}}$

Let us integrate C'₁

$C_1(x)=\lmoustache{({e^{x}\over {e^{x}+1}})dx}=ln\vert{{e^{x}+1}}\vert+C_1$

Now we add these values in our general solution for y and will get

$y_{gen}=(-e^{x}+ln\vert{{e^{x}+1}}\vert+C_1)e^{-{x}}+e^{-{2x}}(ln\vert{{e^{x}+1}}\vert+C_2)$

Finally we get

$y_{gen}=1+e^{-{x}}ln\vert{{e^{x}+1}}\vert+C_1e^{-{x}}+e^{-{2x}}ln\vert{{e^{x}+1}}\vert+C_2e^{-{2x}}$