What is the solution for y^'+4y=3sin(e^{4x}) ?

The solution for y^'+4y=3sin(e^{4x}) is y=-(3cos(e^{4x}))/(4e^{4x)}+(c_{1})/(e^{4x)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'+4y=3sin(e^{4x})

y^{^{'}} + 4 y = 3 sin (e^{4 x})

y = - \frac{3 cos ( e ^{4 x} )}{4 e ^{4 x}} + \frac{c _{1}}{e ^{4 x}}

Solution steps

y^{'} (x) + 4 y = 3 sin (e^{4 x})

Solve linear ODE:

y = - \frac{3 cos ( e ^{4 x} )}{4 e ^{4 x}} + \frac{c _{1}}{e ^{4 x}}

y = - \frac{3 cos ( e ^{4 x} )}{4 e ^{4 x}} + \frac{c _{1}}{e ^{4 x}}

\int \frac{4 x ^{2} - 16}{x} d x

\frac{d}{d x} ((tan (x))^{- 1})

\int sec^{2} (x) tan^{4} (x) d x

x \to \infty lim (e^{\frac{- x}{100}})

x \to - 2 lim (\frac{x + 3}{x + 2})

What is the solution for y^'+4y=3sin(e^{4x}) ?
The solution for y^'+4y=3sin(e^{4x}) is y=-(3cos(e^{4x}))/(4e^{4x)}+(c_{1})/(e^{4x)}