Y” + y = sec3x, y(0) = 1, y'(0) = 1/2 y” + y = 0 m2 + 1 = 0...

y'' + y = sec3x, y(0) = 1, y'(0) = 1/2 y'' + y = 0 m2 + 1 = 0 Solution are -i, i y_c(x) = c_1 cos x + c_2 sin x y_1 = cos x, y_2 = sin x W(y_1, y_2) = |cos x sin x| / |-sin x cos x| = 1 y_p = u_1 y_1 + u_2 y_2 u_1' = - (sin x * sec3x) / 1 = -sin x / cos3x u_2' = (cos x * sec3x) / 1 = 1 / cos2x u_1(x) = -1/2 tan2x u_2(x) = tan x y_p(x) = -1/2 tan2x cos x + tan x sin x y_p(x) = 1/2 tan x sin x ? y(x) = c_1 cos x + c_2 sin x + 1/2 tan x sin x y(0) = 1, y'(0) = 1/2 c_1 = 1, c_2 = 1/2 y(x) = cos x + 1/2 sin x + 1/2 tan x sin x

Transcript

00:01 Hi, here in this question we are given that y double dash plus y equals to 6 square x and y of 0 equals to 1 and y dash of 0 equals to 1 by 2 so here first of all we will find the homogeneous solution so here y double dash plus y equals to 0 can be written as f square plus 1 equals to 0 so here the solution of n is n equals 2 plus n or minus so here the complementary salimation y c of x equals to c1 4x plus x plus x now further let y1 equals to pos x and y2 equals to sine x so here wrong square of the function y1 form of y2 is equal to pos x here sine x and here derivative so minus sine x and cross x so here on expanding this we have this value as square x plus sine square x this is equal to 1 so here we can say that we got the wrong screen of our function wrong screen of y1 comma y 2 equals to 1 so here further we know that the value of the particular solution will be equal to here we have y of p as u1 by 1 plus u2 by 2 so here u1 x x x multiplied with 6 cube x upon 1 so this is equal to minus sine x upon cos x now further similarly we need to calculate value of u2 dash so u2 dash equals to cos x multiplied with sec cube x upon 1 so this is again equal to 1 upon cos x so here again mu 2 x equals to minus 1 by 2 10 x and mu 1 of x equals to 10 x so here using this value we have our particular solution y of p equals to 1 by 2 10 x 10 x plus 10 x plus 10 x multiplied with sinex.

02:37 So here now using the homogeneous solution and complementary solution here we can say that we are also given initial conditions as y of 0 equals to 1 and y dash of 0 equals to 1 by 2.

02:51 So using this we have constant c1 equals to 1 and c2 equals to 1 by 2.

02:58 So here using particular solution homogeneous solution and initial values we can say that we got our equal required solution as y of x equals to 4 6...

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