Question

Find the function of which is the solution of 36y"-48y'-48y=0 with initial conditions y1(0)=1. y1'(0)=0

Answer 1

Answer:

$y=\frac{1}{4} e^{2x}+\frac{3}{4} e^{-\frac{2}{3}x }$

Step-by-step explanation:

Let,  $D=\frac{d}{dx}$

Now, us simplify the given differential equation and write it in terms of D,

$36y''-48y'-48y=0$

or, $3y''-4y'-4y=0$

or, $(3D^2-4D-4)y=0$

We have our auxiliary equation:

$3D^2-4D-4=0$

or, $(3D+2)(D-2)=0$

or, $D=2, - \frac{2}{3}$

Therefore our solution is,

$y=Ae^{2x}+Be^{- \frac{2}{3}x}$

and, $y'=2Ae^{2x}-\frac{2}{3}Be^{\frac{2}{3}x }$

Applying the boundary conditions, we get,

$A+B=1$

$2A-\frac{2}{3}B=0$

Solving them gives us,

$A=\frac{1}{4} ,B=\frac{3}{4}$

Hence,

$y=\frac{1}{4} e^{2x}+\frac{3}{4} e^{-\frac{2}{3}x }$