Answer to Question #178212 in Differential Equations for Ashweta Padhan

Question

Solve -y^''-2y^'+2y = e^-2x

y(0)=1

y(infinity )=0

Answer 1

$-y''-2y'+2y=e^{-2x}$

$y(0)=1$

$y(\infty)=0$

Solution:

$y''+2y'-2y=-e^{-2x}$

Characteristic equation:

$r^2+2r-2=0$ ;

$D=\sqrt{4+4\cdot 2}=2\sqrt3$ ;

$r_1=-1+\sqrt3$ ; $r_2=-1-\sqrt3$ .

The general solution of the homogeneous differential equation:

$y_0=C_1e^{(-1+\sqrt3)x}+C_2e^{(-1-\sqrt3)x}$

Let's find a partial solution of the non-homogeneous equation in the form

$y_p=Ae^{-2x}$

$y_p'=-2Ae^{-2x}$

$y_p''=4Ae^{-2x}$

Substitute $y_p$, $y_p'$, $y_p''$ into the origin differential equation:

$4Ae^{-2x}-4Ae^{-2x}-2Ae^{-2x}=-e^{-2x}$

$A=\frac12$

The general solution of the non-homogeneous differential equation:

$y=y_0+y_p=C_1e^{(-1+\sqrt3)x}+C_2e^{(-1-\sqrt3)x}+\frac12e^{-2x}$

Let's apply conditions:

$y(0)=C_1+C_2+\frac12=1$

$y(\infty)=0\implies$ $C_1=0$ (because if $C_1\neq0$ then $y(\infty)\to\infty$ )

$C_2=\frac12$

Answer: $y=\frac12e^{(-1-\sqrt3)x}+\frac12e^{-2x}$ .