Question

solution for y''+y=sec x tan^2x by variation of parameters

Answer 1

since there is no table readily available for the particular case, we can go the long route or the short route.

first step is to find the homogenous part. lets assume some sort of e^rx will suffice.

y = e^rx
y'= rer^x
y'' = r^2 e^rx

now lets put these into the diffyQ to see what we can come up with.

r^2 e^rx + e^rx = 0

e^rx (r^2+1) = 0 ; this is only 0 when the quadratic goes zero

r^2 + 1 = 0

r^2 = -1

r = +- sqrt(-1) = +-i

\[y=e^0(c_1cos(x)+c_2sin(x)) + y_p\]
\[y=c_1cos(x)+c_2sin(x) + y_p\]

Answer 2

now we have all the parts to form the wronslian with:\[\begin{vmatrix}W_1&W&W_2\\cos(x)&0&sin(x)\\-sin(x)&sec(x)tan^2(x)&cos(x)\end{vmatrix}\]

if we take somethe pseudo, determinant of the matrix we get:

W1 = -tan^3(x)
W = 1
W2 = tan^2(x)
\[y_p=cos(x)\int -tan^3(x)dx+sin(x)\int tan^2(x)dx\]

Answer 3

this is actually the short route :)

Answer 4

the long route i believe would be to assume c1 and c2 are functions of x such that:

\[y_p=A(x)cos(x)+B(x)sin(x)\]
\[{y_p}'=-A(x)sin(x)+A'(x)cos(x)+B(x)cos(x)+B'(x)sin(x)\]\[assume:\ A'cos(x)+B'sin(x)=0\]

\[{y_p}'=-A(x)sin(x)+B(x)cos(x)+0\]
\[\begin{array}l {y_p}''&=-A(x)cos(x)-A'(x)sin(x)-B(x)sin(x)+B'(x)cos(x)\\+y_p &=A(x)cos(x)+B(x)cos(x)\\---&-------------------------\\
sec(x)tan^2(x)&=-A'(x)sin(x)+B'(x)cos(x)\end{array} \]

this is a system of 2 equations in 2 unknowns

\[A'cos(x)+B'sin(x)=0\]
\[-A'sin(x)+B'cos(x)=sec(x)tan^2(x)\]