Question

Solve the initial value problem y"-3y'-2y=0; y(0)=1, y(3)=0

Answer 1

The auxiliary equation is $m^2-3m-2=0$

$m= \dfrac{3\pm \sqrt{9-4(1)(-2)}}{2} = \dfrac{3 \pm \sqrt{17}}{2}$

$m = \dfrac{3+\sqrt{17}}{2}$ Or $m = \dfrac{3-\sqrt{17}}{2}$

$m = 3.56$ or $m = -0.56$

Hence the general solution is $y = Ae^{3.56x} + Be^{-0.56x}$

So $1= y(0) = A+B .... (*)$

And $0= y(3)= 43477.55A+0.19B .... (**)$

So $B = \dfrac{-43477.55A}{0.19} = - 228829.21A$

From equation (*) 1=A−228829.21A=−228828.21A

So A =−4.37×10$^{-6}$

And B = 0.99998

So the solution of the IVP is y = $-4.37 \times 10^{-6}e^{3.56x} + 0.99998e^{-0.56x}$

NOTE: Although what I have written down is the logical solution for the prescribed question. But I think there's a mistake in the question especially with respect to the boundary condition.