Answer:
[tex]y=\frac{1}{4} e^{2x}+\frac{3}{4} e^{-\frac{2}{3}x }[/tex]
Step-by-step explanation:
Let, [tex]D=\frac{d}{dx}[/tex]
Now, us simplify the given differential equation and write it in terms of D,
[tex]36y''-48y'-48y=0[/tex]
or, [tex]3y''-4y'-4y=0[/tex]
or, [tex](3D^2-4D-4)y=0[/tex]
We have our auxiliary equation:
[tex]3D^2-4D-4=0[/tex]
or, [tex](3D+2)(D-2)=0[/tex]
or, [tex]D=2, - \frac{2}{3}[/tex]
Therefore our solution is,
[tex]y=Ae^{2x}+Be^{- \frac{2}{3}x}[/tex]
and, [tex]y'=2Ae^{2x}-\frac{2}{3}Be^{\frac{2}{3}x }[/tex]
Applying the boundary conditions, we get,
[tex]A+B=1[/tex]
[tex]2A-\frac{2}{3}B=0[/tex]
Solving them gives us,
[tex]A=\frac{1}{4} ,B=\frac{3}{4}[/tex]
Hence,
[tex]y=\frac{1}{4} e^{2x}+\frac{3}{4} e^{-\frac{2}{3}x }[/tex]
