What is the solution for y^'+ycos(x)=sin(x)cos(x) ?

The solution for y^'+ycos(x)=sin(x)cos(x) is y=sin(x)-1+(c_{1})/(e^{sin(x))}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'+ycos(x)=sin(x)cos(x)

y^{^{'}} + y cos (x) = sin (x) cos (x)

y = sin (x) - 1 + \frac{c _{1}}{e ^{s i n (x)}}

Solution steps

y^{'} (x) + y cos (x) = sin (x) cos (x)

Solve linear ODE:

y = sin (x) - 1 + \frac{c _{1}}{e ^{s i n (x)}}

y = sin (x) - 1 + \frac{c _{1}}{e ^{s i n (x)}}

f (x) = (ln (x))^{6}, at x = 4

x y^{^{'}} + 2 y = - 3 cos (x)

x^{2} + 6

f (x) = ln (x) + 1

x^{3} - x^{2}

What is the solution for y^'+ycos(x)=sin(x)cos(x) ?
The solution for y^'+ycos(x)=sin(x)cos(x) is y=sin(x)-1+(c_{1})/(e^{sin(x))}