[SOLVED] Differential

**ronaldo_07** · March 6th 2009, 01:48 PM

Find the general solution of the second-order, linear, inhomogeneous
differential equation:

y′′ + 3y′ + 2y = $\frac{1}{(1 + e^x)^2}$

Please help solve this equation so far I have done:

z^2 +3z+2=0
(z+2)(z+1)=0
z=-2,-1

**Chris L T521** · March 6th 2009, 03:46 PM

Originally Posted by ronaldo_07

Find the general solution of the second-order, linear, inhomogeneous
differential equation:

y′′ + 3y′ + 2y = $\frac{1}{(1 + e^x)^2}$

Please help solve this equation so far I have done:

z^2 +3z+2=0
(z+2)(z+1)=0
z=-2,-1

Looks good up to here.

The homogeneous solution is $y_c=c_1e^{-2x}+c_2e^{-x}$

Here, I would consider using Variation of Parameters.

Let $y_p=u_1\!\left(x\right)e^{-2x}+u_2\!\left(x\right)e^{-x}$ .

Thus, we need to evaluate three different Wronskians.

$W_1=\begin{vmatrix}0 & e^{-x}\\ \frac{1}{\left(1+e^x\right)^2} &-e^{-x}\end{vmatrix}=-\frac{e^{-x}}{\left(1+e^x\right)^2}$

$W_2=\begin{vmatrix}e^{-2x} & 0\\ -2e^{-2x} & \frac{1}{\left(1+e^x\right)^2} \end{vmatrix}=\frac{e^{-2x}}{\left(1+e^x\right)^2}$

$W=\begin{vmatrix}e^{-2x} & e^{-x} \\ -2e^{-2x} & -e^{-x}\end{vmatrix}=-e^{-3x}+2e^{-3x}=e^{-3x}$

Thus, $u_1^{\prime}\!\left(x\right)=\frac{W_1}{W}=-\frac{e^{2x}}{\left(1+e^x\right)^2}$ and $u_2^{\prime}\!\left(x\right)=\frac{W_2}{W}=\frac{e ^x}{\left(1+e^x\right)^2}$

Thus,

$u_1\!\left(x\right)=-\int \frac{e^{2x}}{\left(1+e^x\right)^2}\xrightarrow{z= 1+e^x}{}-\int\frac{z-1}{z^2}\,dz=-\ln\left|z\right|-\frac{1}{z}\implies -\ln\left|1+e^x\right|-\frac{1}{1+e^{x}}$

and

$u_2\!\left(x\right)=\int\frac{e^x}{\left(1+e^x\rig ht)^2}\,dx\xrightarrow{z=1+e^x}{}\int\frac{\,dz}{z ^2}=-\frac{1}{z}\implies -\frac{1}{1+e^x}$

Thus, our solution is

$y=c_1e^{-2x}+c_2e^{-x}-\left(\ln\left|1+e^x\right|-\frac{1}{1+e^x}\right)e^{-2x}+\left(-\frac{1}{1+e^{-x}}\right)e^{-x}$

$=c_1e^{-2x}+c_2e^{-x}-\frac{\left(\ln\left(1+e^x\right)e^{-2x}\right)\left(1+e^x\right)+e^{-2x}+e^{-x}}{1+e^x}$

$=\color{red}\boxed{c_1e^{-2x}+c_2e^{-x}-\frac{e^{-2x}\ln\left(1+e^x\right)+e^{-x}\ln\left(1+e^x\right)+e^{-2x}+e^{-x}}{1+e^x}}$

Does this make sense?

**ronaldo_07** · March 6th 2009, 03:51 PM

Could you explain Thus, we need to evaluate three different Wronskians the W W1 W2. How this is calculated please. Thanks.

**Chris L T521** · March 6th 2009, 04:08 PM

Originally Posted by ronaldo_07

Could you explain Thus, we need to evaluate three different Wronskians the W W1 W2. How this is calculated please. Thanks.

See my explanation here.

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