Question

(D^2+2D+1)y=(e^x-1)^-2  solve by reduction order

Answer 1

(D²+2D+1)y=y"+2y'+y=(e^x-1)^-2.

y"+2y'+y=(y"+y')+(y'+y). Let z=y'+y then (D²+2D+1)y=z'+z=(e^x-1)^-2.

This reduces the 2nd order DE to a 1st order DE.

We can use the integrating factor e^x to reduce this further:

d(e^xz)=e^x(e^x-1)^-2, so e^xz=∫e^x(e^x-1)^-2dx.

Let u=e^x-1, then du=e^xdx, and:

e^xz=∫u^-2du=C-u^-1=C-(e^x-1)^-1, z=Ce^-x-e^-x(e^x-1)^-1, where C is the arbitrary constant of integration.

z=y'+y=Ce^-x-e^-x(e^x-1)^-1.

Using the same integrating factor e^x:

d(e^xy)=C-(e^x-1)^-1, e^xy=∫(C-(e^x-1)^-1)dx=Cx-∫(e^x-1)^-1dx.

du=e^xdx=(u+1)dx, e^xy=Cx-∫u^-1(u+1)^-1du.

u^-1(u+1)^-1≡A/(u+1)+B/u where A and B are constants to be found (to create partial fractions).

Au+Bu+B=1, so B=1 and A=-1.

e^xy=Cx-∫(-1/(u+1)+1/u)du=Cx+ln|u+1|-ln|u|+D, where D is another constant of integration.

u+1=e^x, so ln|u+1|=ln|e^x|=x.

e^xy=Cx+x-ln|e^x-1|+D, y=C₁xe^-x-e^-xln|e^x-1|+De^-x. Cx+x has been replaced by C₁=C+1.

This result needs to be checked...

To simplify the checking process we use the fact that (D²+2D+1)y=0 has the characteristic solution y_c=ae^-x+bxe^-x where a and b are constants. In the solution D=a, C₁=b. The particular solution is the term -e^-xln|e^x-1|, so we only need to apply the DE to this term to prove the solution.

y_p=-e^-xln|e^x-1|, y_p'=e^-xln|e^x-1|-(e^x-1)^-1,

+2y_p'=2e^-xln|e^x-1|-2(e^x-1)^-1,

+y_p"=-e^-xln|e^x-1|+(e^x-1)^-1+e^x(e^x-1)^-2. The red terms cancel out.

Add the terms together:

(e^x-1)^-1+e^x(e^x-1)^-2-2(e^x-1)^-1=e^x(e^x-1)^-2-(e^x-1)^-1=

(e^x-1)^-1(e^x(e^x-1)^-1-1)=(e^x-1)^-1(e^x-e^x+1)(e^x-1)^-1=(e^x-1)^-2.

This confirms the general solution y_c+y_p=y=ae^-x+bxe^-x-e^-xln|e^x-1|.