What is the solution for y^{'''}+y^'=sec^2(x) ?

The solution for y^{'''}+y^'=sec^2(x) is y=c_{1}+c_{2}cos(x)+c_{3}sin(x)-cos(x)ln(sec(x)+tan(x))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{'''}+y^'=sec^2(x)

y^{^{'''}} + y^{^{'}} = sec^{2} (x)

y = c_{1} + c_{2} cos (x) + c_{3} sin (x) - cos (x) ln (sec (x) + tan (x))

Solution steps

y^{'''} (x) + y^{'} (x) = sec^{2} (x)

Solve linear ODE:

y = c_{1} + c_{2} cos (x) + c_{3} sin (x) - cos (x) ln (sec (x) + tan (x))

y = c_{1} + c_{2} cos (x) + c_{3} sin (x) - cos (x) ln (sec (x) + tan (x))

What is the solution for y^{'''}+y^'=sec^2(x) ?
The solution for y^{'''}+y^'=sec^2(x) is y=c_{1}+c_{2}cos(x)+c_{3}sin(x)-cos(x)ln(sec(x)+tan(x))