Question

A system is described by the inhomogeneous partial differential equation Sxx + 27,+/- = 41e-x/2sin( 3x) with boundary conditions f(t,0) =0, f (1,8) = 0 and f(0,x) =0, where subscripts represent partial derivatives. Using Duhamel's Principle, the auxiliary solution is w(1,x;s) = 2.se – 37518 37 1/8 = x/2sin( 3x). What is the solution (1,8) ? OA) 16 2 = x/2( 1– e – 37418) + sin( 3x) 37 B) 32 -e-x/24 – 8+8 e37418 – 37t) sin( 3x) 1369 OC) 16 1369 -x/2( - 8+8 e371/8 – 371) sin( 3x) OD) 8 -e-x/27 - 4e – 371/8 + 4e 371/8 – 376 e - 37118) sin( 3x) 1369

Answer 1

so here we have given that an inhomogeneous equation and we have to find the solution f(t,x) with the help of

DUHAMEL PRINCIPLE WHICH relates the solution of the given equation with auxiliary solution

and we have given already the Auxiliary equation .so here we dont need to find it

we just find the solution f(t,x) with the help of this formula given below

Duhamel's Principle: The solution of (*) is given by u(z, t) = (,t;s)ds. Thus, we have

so f(t,x)=$\int_{0}^{t}$ w(t,x,s) ds

=2e^37t/8-x/2 sin3x  $\int_{0}^{t}$(s.e^-37s/8)ds

now we solve with help of integration by parts which is stated below

${\displaystyle {\begin{aligned}\int _{a}^{b}u(x)v'(x)dx&={\Big [}u(x)v(x){\Big ]}_{a}^{b}-\int _{a}^{b}u'(x)v(x)dx\\[6pt]&=u(b)v(b)-u(a)v(a)-\int _{a}^{b}u'(x)v(x)dx.\end{aligned}}}$

=2 e^37t/8-x/2 sin3x [s{e^-37s/8 (-8/37)}+ 8/37$\int_{0}^{t}$ e^-37s/8 ds]  

=2 e^37t/8-x/2 sin3x [s{e^-37s/8(-8/37)} + 8/37 {(-8/37)e^-37s/8}]

taking limit 0 to t in complete integral where you can see the closed bracket [ ]

=2.e^37t/8-x/2 sin3x [(-8/37t.e^-37t/8)-(64/1369.e^-37t/8+ 64/1369)]

= e^-x/2 sin3x [ -16/37 t - 128/1369 + 128/1369 e^37t/8]

= 16/1369 .e^-x/2 sin3x [-37t-8+8.e^37t/8]

so option (C) is the correct answer