Solve the following system of ODE
(48D+4)y - (2D+1)z = e
-x
(D+8)y - 3z = 5e
-x
; D =d/dx
{48dxdy+4y−2dxdz−z=e−xdxdy+8y−3z=5e−x
Solution:
From second equation:
z=31(dxdy+8y−5e−x) (*)
dxdz=31(dx2d2y+8dxdy+5e−x)
Substitute z and dxdz into the first equation:
48dxdy+4y−32(dx2d2y+8dxdy+5e−x)−31(dxdy+8y−5e−x)=e−x
−32dx2d2y+3127dxdy+34y=38e−x
2dx2d2y−127dxdy−4y=−8e−x (**)
First find the general solution of the corresponding homogeneous equation:
2dx2d2y0−127dxdy0−4y0=0
Let's compose and solve the characteristic equation:
2λ2−127λ−4=0
λ1=4127−16161 , λ2=4127+16161 .
y0=C1eλ1x+C2eλ2x=C1e4127−16161x+C2e4127+16161x .
Find a typical (specific) solution of the non-homogeneous equation in the form yp=Ae−x
Derivatives of the solution: dxdyp=−Ae−x , dx2d2yp=Ae−x
Substitute those "solutions" into the (**):
2Ae−x+127Ae−x−4Ae−x=−8e−x
2A+127A−4A=−8
A=−1258
yp=−1258e−x .
Add the typical and the complementary solutions to get the complete solution:
y=y0+yp=C1e4127−16161x+C2e4127+16161x−1258e−x .
To find z substitute y into the (*):
z=31(C14127−16161e4127−16161x+C24127+16161e4127+16161x+1258e−x+
8(C1e4127−16161x+C2e4127+16161x−1258e−x)−5e−x)=
C112159−16161e4127−16161x+C212159+16161e4127+16161x−125227e−x
Answer: {y=C1e4127−16161x+C2e4127+16161x−1258e−xz=C112159−16161e4127−16161x+C212159+16161e4127+16161x−125227e−x
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