Solve the given differential equation by undetermined coeffi
Solve the given differential equation by undetermined coefficients y\'\'-8y\'+16y=24x+2
Solution
Given that
y\'\' - 8y\' + 16y = 24x + 2
d2y/dx2 - 8dy/dx + 16y = 24x + 2................................1
The D-operator form is ,
( D2 - 8D + 16 ) y = 24x + 2
The auxialary equation is,
m2 - 8m + 16 = 0
m2 - 4m - 4m + 16 = 0
m ( m - 4 ) - 4 ( m - 4 ) = 0
( m - 4 ) ( m - 4 ) = 0
m = 4 , 4
m1 = 4 , m2 = 4
If the roots are real and equal then the complementary function is ,
yc = c1 em1x + c2xem2x
yc = c1e4x + c2xe4x
For a non homogeneous term 24x + 2 assume perticular solution is,
yp = Ax + B
Substitute y = Ax + B in equation 1
d2y/dx2 - 8dy/dx + 16y = 24x + 2
d2/dx2 ( Ax + B ) - 8d/dx( Ax + B ) + 16 ( Ax + B ) = 24x + 2..............................2
d/dx( Ax + B ) = A.d/dx(x) + d/dx( B ) = A + 0 = A [ since,d/dx( x ) = 1]
d2/dx2( Ax + B ) = d/dx( A) = 0 [ since , d/dx( constant ) = 0 ]
From equation 2
d2/dx2 ( Ax + B ) - 8d/dx( Ax + B ) + 16 ( Ax + B ) = 24x + 2
0 - 8A + 16Ax + 16B = 24x + 2
16Ax + ( -8A + 16B ) = 24x + 2
Equating the coefficients
16A = 24
A = 24 / 16
A = 3 / 2
-8A + 16B = 2
-8(3/2) + 16B = 2
-12 + 16B = 2
16B = 2 + 12
16B = 14
B = 14 / 16
B = 7 / 8
Hence,
Perticular solution is ,
yp = Ax + B
yp = ( 3/2 )x + ( 7/8 )
The general solution is , y(x) = yc + yp
y(x) = c1e4x + c2xe4x + ( 3/2 )x + ( 7/8 )
Therefore,
The general solution is , y(x) = c1e4x + c2xe4x + ( 3/2 )x + ( 7/8 )
