What is the solution for xy^'+4y=-25xsin(x^5) ?

The solution for xy^'+4y=-25xsin(x^5) is y=(5cos(x^5))/(x^4)+(c_{1})/(x^4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

xy^'+4y=-25xsin(x^5)

x y^{^{'}} + 4 y = - 25 x sin (x^{5})

y = \frac{5 cos ( x ^{5} )}{x ^{4}} + \frac{c _{1}}{x ^{4}}

Solution steps

x y^{'} (x) + 4 y = - 25 x sin (x^{5})

Solve linear ODE:

y = \frac{5 cos ( x ^{5} )}{x ^{4}} + \frac{c _{1}}{x ^{4}}

y = \frac{5 cos ( x ^{5} )}{x ^{4}} + \frac{c _{1}}{x ^{4}}

\frac{d}{d x} (\frac{3}{7 x ^{4} - 5 x ^{2} + 1})

\frac{d y}{d x} = tan (4 x)

\int 16 - t^{2} d t

\frac{\partial}{\partial y} (1 - x - y)

x \to 0.1 lim (\frac{e ^{x} - 1}{x})

What is the solution for xy^'+4y=-25xsin(x^5) ?
The solution for xy^'+4y=-25xsin(x^5) is y=(5cos(x^5))/(x^4)+(c_{1})/(x^4)