What is the solution for y^'+8(tan(8x))*y=-9cos(8x) ?

The solution for y^'+8(tan(8x))*y=-9cos(8x) is y=-(9x)/(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=-9cos(8x)

y^{^{'}} + 8 (tan (8 x)) \cdot y = - 9 cos (8 x)

y = - \frac{9 x}{sec ( 8 x )} + \frac{c _{1}}{sec ( 8 x )}

Solution steps

y^{'} (x) + 8 (tan (8 x)) \cdot y = - 9 cos (8 x)

Solve linear ODE:

y = - \frac{9 x}{sec ( 8 x )} + \frac{c _{1}}{sec ( 8 x )}

y = - \frac{9 x}{sec ( 8 x )} + \frac{c _{1}}{sec ( 8 x )}

(0.5) (7.9)

\int 9 e^{5 x + e^{5 x}} d x

\int x (4 x - 1)^{4} d x

\int \frac{1}{4 + ( u + 3 )} d u

y = ln (8 x)

What is the solution for y^'+8(tan(8x))*y=-9cos(8x) ?
The solution for y^'+8(tan(8x))*y=-9cos(8x) is y=-(9x)/(sec(8x))+(c_{1})/(sec(8x))