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MTH401 Idea of Solution Feb 2015
#1
[Image: example.png]

[Image: remainingexample.png]

For a second-order ordinary differential equation,
y^('')+p(x)y^'+q(x)y=g(x).
(1)
Assume that linearly independent solutions y_1(x) and y_2(x) are known to the homogeneous equation
y^('')+p(x)y^'+q(x)y=0,
(2)
and seek v_1(x) and v_2(x) such that
y^* = v_1y_1+v_2y_2
(3)
y^('*) = (v_1^'y_1+v_2^'y_2)+(v_1y_1^'+v_2y_2^').
(4)
Now, impose the additional condition that
v_1^'y_1+v_2^'y_2=0
(5)
so that
y^('*)(x) = v_1y_1^'+v_2y_2^'
(6)
y^(''*)(x) = v_1^'y_1^'+v_2^'y_2^'+v_1y_1^('')+v_2y_2^('').
(7)
Plug y^*, y^*^', and y^*^('') back into the original equation to obtain
v_1(y_1^('')+py_1^'+qy_1)+v_2(y_2^('')+py_2^'+qy_2)+v_1^'y_1^'+v_2^'y_2^'=g(x),
(8)
which simplifies to
v_1^'y_1^'+v_2^'y_2^'=g(x).
(9)
Combing equations (◇) and (9) and simultaneously solving for v_1^' and v_2^' then gives
v_1^' = -(y_2g(x))/(W(x))
(10)
v_2^' = (y_1g(x))/(W(x)),
(11)
where
W(y_1,y_2)=W(x)=y_1y_2^'-y_2y_1^'
(12)
is the Wronskian, which is a function of x only, so these can be integrated directly to obtain
v_1 = -int(y_2g(x))/(W(x))dx
(13)
v_2 = int(y_1g(x))/(W(x))dx,
(14)
which can be plugged in to give the particular solution
y^*=v_1y_1+v_2y_2.
(15)
Generalizing to an nth degree ODE, let y_1, ..., y_n be the solutions to the homogeneous ODE and let v_1^'(x), ..., v_n^'(x) be chosen such that
{y_1v_1^'+y_2v_2^'+...+y_nv_n^'=0; y_1^'v_1^'+y_2^'v_2^'+...+y_n^'v_n^'=0; |; y_1^((n-1))v_1^'+y_2^((n-1))v_2^'+...+y_n^((n-1))v_n^'=g(x).
(16)
and the particular solution is then
y^*(x)=v_1(x)y_1(x)+...+v_n(x)y_n(x),
[Image: 1z174mb.gif]
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