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

The solution for y^'+y*cos(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 \cdot cos (x) = sin (x) \cdot cos (x)

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

Solution steps

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

\int \frac{e ^{x} + sec ^{2} ( x )}{e ^{x} + tan ( x )} d x

\frac{s - 4}{( s ^{2} + 5 ) ^{2}}

\int_{0}^{4} x x^{2} + 9 d x

\int \frac{1}{x ^{4} x ^{2} + 3} d x

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