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Reduction of Order - (3.2)
Consider the 2nd-order linear differential equation
a 2 Ÿx y UU a 1 Ÿx y U a 0 Ÿx y 0
The general solution of the equation y C 1 y 1 C 2 y 2 where y 1 , y 2 is a fundamental set of solutions, i.e. they
are solutions of the equation and are linearly independent. If we know y 1 , how can we find y 2 ?
Consider the 2nd-order linear differential equation in standard form
y UU PŸx y U QŸx y 0
Let y 2 y 1 u. Then y U2 y U1 u y 1 u U and y UU2 y UU1 u y U1 u U y U1 u U y 1 u UU y UU1 u 2y U1 u U y 1 u UU .
y UU PŸx y U QŸx y y UU1 u 2y U1 u U y 1 u UU PŸx Ÿy U1 u y 1 u U QŸx y 1 u
Ÿy UU1 PŸx y U1 QŸx y 1 u 2y U1 u U y 1 u UU PŸx y 1 u U
0 y 1 u UU Ÿ2y U1 PŸx y 1 u U 0
u UU zU 2y U1
U
U
U
UU
y 1 PŸx u 0, let z u . Then z u
2y U1
y 1 PŸx z 0
"
It is 1st order linear differential equation in z.
It is also separable.
2y U1
linear in z :
y 1 PŸx z 0 :
2y U
i. hŸx ; y 11 PŸx dx 2 ln|y 1 | ; PŸx dx
Solve z U integrating factor: e hŸx e
ii. kŸx 0
iii. z e
;
"2 ln|y 1 |" PŸx dx
;
2 ln|y 1 | PŸx dx
y 21 e
; PŸx dx
"; PŸx dx
ŸC 0 C 12 e
y1
Steps to Compute y 2 for a Given y 1 :
1. Compute hŸx ; PŸx dx.
2. Compute z 12 e "hŸx .
y1
3. Find u ; zŸx dx.
4. y 2 u y 1 .
Example Ÿ1 Show that y 1 x 2 cosŸln x is a solution of x 2 y UU " 3xy U 5y 0. Ÿ2 Find the general
solution of the differential equation.
Ÿ1 Verify:
y U1 2x cosŸln x " x 2 sinŸln x 1x xŸ2 cosŸln x " sinŸln x ,
y UU1 2 cosŸln x " sinŸln x x "2 sinŸln x 1x " cosŸln x 1x cosŸln x " 3 sinŸln x x 2 y UU " 3xy U 5y x 2 ŸcosŸln x " 3 sinŸln x " 3xŸx Ÿ2 cosŸln x " sinŸln x 5x 2 cosŸln x Ÿx 2 " 6x 2 5x 2 cosŸln x Ÿ"3x 2 3x 2 sinŸln x 0
Ÿ2 Rewrite the equation: xy UU " 3x y U 5x y 0: PŸx " 3x
1
1. hŸx ; " 3x dx "3 ln|x|
1
1
1
e 3|x| 4
2. z 4
1x sec 2 ŸlnŸx Ÿx 3 x cos 2 ŸlnŸx x cos 2 ŸlnŸx x cos 2 ŸlnŸx 3. u ; zdx ; 1x sec 2 ŸlnŸx dx tanŸlnŸx 4. y 2 ŸtanŸln x Ÿx 2 cosŸln x x 2 sinŸln x 5. The general solution: y C 1 x 2 cosŸln x C 2 x 2 sinŸln x 2