Answer to Question #195581 in Differential Equations for Ashweta Padhan

Question

Solve the boundary value problem

y''-y=0

With boundary condition

y(0)=0, y(2)=3.6268

Answer 1

Given,

y''-y=0

$(D^2-1)y=0$

Auxiliary equation is

$m^2-1=0\\(m-1)(m+1)=0 \\m=1,-1$

Solution of the equation is-

$y=c_1 e^x+c_2 e^{-x}$

As $y(0)=0\Rightarrow 0=c_1+c_2~~~~-(1)$

Also, $y(2)=3.6268\Rightarrow 3.6268=c_1 e^2+c_2e^{-2}$

$7.389c_1+0.1353c_2=3.6268~~~~~~-(2)$

Solving equation (1) and (2) and we get-

$c_1=0.5,c_2=-0.5$

So Solution is-

$y=0.5e^x-0.5e^{-x}$