$2\frac{d^2y}{dx^2} + 7\frac{dy}{dx} -y =0$ where y(0)=1 and y'(0)=3

Question

I need help with this slightly tedious question

the got the solution of the auxiliary equation to be $s=\frac{-7+\sqrt{57} }{2}$ and $t=\frac{-7-\sqrt{57} }{2}$ thus the general solution $y=Ae^{sx}+Be^{tx}$.

Given the initial conditions I obtained the two simultaneous equations $1=A+B$ and $\frac{20}{\sqrt{57} }=A-B$ which gave me $A=1.824532357$ and $B=-0.824532357$

so when x=2 my y value was 3.16 (2dp) hich was incorrect but I don't see where I went wrong.

Can anyone PLEASE confirm what answer they get to 2dp (Note: you must be as accurate as possible)

Many thanks

Answer 1

As L.F. stated, you have an issue with your denominator from your auxiliary equation.

You should have, $2m^2 + 7m -1 = 0$, yielding:

$~m = \frac{1}{4} (-\sqrt{57}-7) $ and $m = \frac{1}{4} (\sqrt{57}-7)$

From this and the IC's, we arrive at:

$$\displaystyle \large y(x) = \frac{1}{6} e^{-\frac{1}{4} (7+\sqrt{57}) x} \left((3+\sqrt{57}) e^{\frac{\sqrt{57} x}{2}}+3-\sqrt{57}\right)$$

At $x = 2$, we get:

$$y(2) = \displaystyle \frac{1}{6} e^{\frac{1}{2} (-7-\sqrt{57})} \left(3-\sqrt{57}+(3+\sqrt{57}) e^{\sqrt{57}}\right) \approx 2.3141465430$$