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idea to solve 1st question
what will be the complementary solution of the given equation ??
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),
kuch samaj ni a rha. plz Naseer dobara post kry shi kr k.
solution of second question
bro is ma ap na singular regular or irragular points to btya nhi plz anyone tell what is singular regular or irragular points?
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