What is the solution for y^'+8(tan(8x))y=-2cos(8x) ?

The solution for y^'+8(tan(8x))y=-2cos(8x) is y=-(2x)/(sec(8x))+(c_{1})/(sec(8x))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'+8(tan(8x))y=-2cos(8x)

y^{^{'}} + 8 (tan (8 x)) y = - 2 cos (8 x)

y = - \frac{2 x}{sec ( 8 x )} + \frac{c _{1}}{sec ( 8 x )}

Solution steps

y^{'} (x) + 8 (tan (8 x)) y = - 2 cos (8 x)

Solve linear ODE:

y = - \frac{2 x}{sec ( 8 x )} + \frac{c _{1}}{sec ( 8 x )}

y = - \frac{2 x}{sec ( 8 x )} + \frac{c _{1}}{sec ( 8 x )}

\frac{( t ^{6} )}{( t ^{3} + 1 ) ^{2}}

\int (1 - sin (θ))^{2} d θ

\int cos (- 1 (x)) d x

\int 2 x (x^{2} + 4)^{5} d x

y = \frac{6 x}{e ^{x}}

What is the solution for y^'+8(tan(8x))y=-2cos(8x) ?
The solution for y^'+8(tan(8x))y=-2cos(8x) is y=-(2x)/(sec(8x))+(c_{1})/(sec(8x))