Question

Solve the initial-value problem:
\[\huge 4y'' - 4y' + y = 0 \]
\[\huge y(0) = 1 , y'(0) = -1.5\]

Answer 1

The auxillary equation would be:
\[4r^2 - 4r + 1 = 0\]
The root is
r = 1/2

\[\large y = c_1e^{1/2x} + c_2 xe^{1/2x}\]

Answer 2

What a curious high school kid.

Answer 3

Characteristic polynomial is \(\large 4r^2-4r+1 = 0\implies (2r-1)^2=0\implies r= \frac{1}{2}\) with multiplicity 2. Thus, \(\large y = c_1e^{x/2} + c_2xe^{x/2}\). Now, \(\large y^{\prime} = \frac{1}{2}c_1e^{x/2} + c_2(e^{x/2}+\frac{1}{2}xe^{x/2})\). Applying the initial conditions leaves us with \[\large \left\{\begin{aligned}c_1 &= 1\\ \frac{1}{2}c_1 + c_2 &= -\frac{3}{2}\end{aligned}\right.\implies c_1=1\quad\text{and}\quad c_2 = -2\]Therefore \(\large y=e^{x/2} -2xe^{x/2}\).

Answer 4

\[y'(x) = (1/2)e^{1/2x} + c_2(e^{x/2} + 1/2xe^{x/2})\]
\[1= c_1\]
\[-1.5 = \frac{ 1 }{ 2 }c_1 e^{x/2} + c_2 (e^{x/2} + 1/2x e^{x/2})\]

Answer 5

yeah, i was learning some calc 3 today lol

Answer 6

I'm curious though, how did @ChristopherToni you evaluate c_2 so fast? :o

Answer 7

plug in 0 into \(\large y^{\prime}\) and it should simplify to \(\large\frac{1}{2}c_1 + c_2 = -\frac{3}{2}\).

Answer 8

and since \(\large c_1=1\) then we have \(\large \frac{1}{2} + c_2 = -\frac{3}{2} \implies c_2=-\frac{4}{2}=-2\).

Answer 9

\[-1.5 = 1/2 c_1 + c_2 (1)\]
\[-1.5 = 1/2 + c_2\]
\[c_2 = -2\]

Answer 10

yeah got it now xD, brain is slow, i need sleep lol

Answer 11

\[y = c_1e^{1/2x}+c_2xe^{1/2x}\]
pluggin em in
c_ 1 = 1
c_2 = -2
\[y = e^{\frac{ 1 }{ 2 }x} - 2xe^{\frac{ 1 }{ 2 }x}\]

Answer 12

Good job. :D

Answer 13

ty, learned this topic today, just seeing if i still remember xD