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Replies to This Discussion

Lets srat the disscusion anoyone help

idea to solve 1st question


what will be the complementary solution of the given equation ??

For a second-order ordinary differential equation,

Assume that linearly independent solutions y_1(x) and y_2(x) are known to the homogeneous equation

and seek v_1(x) and v_2(x) such that

y^* = v_1y_1+v_2y_2
y^('*) = (v_1^'y_1+v_2^'y_2)+(v_1y_1^'+v_2y_2^').
Now, impose the additional condition that

so that

y^('*)(x) = v_1y_1^'+v_2y_2^'
y^(''*)(x) = v_1^'y_1^'+v_2^'y_2^'+v_1y_1^('')+v_2y_2^('').
Plug y^*, y^*^', and y^*^('') back into the original equation to obtain

which simplifies to

Combing equations (◇) and (9) and simultaneously solving for v_1^' and v_2^' then gives

v_1^' = -(y_2g(x))/(W(x))
v_2^' = (y_1g(x))/(W(x)),

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
v_2 = int(y_1g(x))/(W(x))dx,
which can be plugged in to give the particular solution

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).
and the particular solution is then


kuch samaj ni a rha. plz Naseer dobara post kry shi kr k.

plz koi question 2 ka b idea dada us ka single regular and irrgular points kesy niklta h?

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?

(x-2)and (x+3)both are regular singular points


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