What is the solution for xy^'+2y=3cos(x) ?

The solution for xy^'+2y=3cos(x) is y=(3(xsin(x)+cos(x)))/(x^2)+(c_{1})/(x^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

xy^'+2y=3cos(x)

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

y = \frac{3 ( x sin ( x ) + cos ( x ) )}{x ^{2}} + \frac{c _{1}}{x ^{2}}

Solution steps

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

Solve linear ODE:

y = \frac{3 ( x sin ( x ) + cos ( x ) )}{x ^{2}} + \frac{c _{1}}{x ^{2}}

y = \frac{3 ( x sin ( x ) + cos ( x ) )}{x ^{2}} + \frac{c _{1}}{x ^{2}}

\int cos^{2} (2 x) sin^{3} (2 x) d x

h (x) = 4^{x} - 6

g (x) = 3 x^{2} cos (x) + 8 x

\int \frac{x}{( x ^{2} + 9 )} d x

x - 5

What is the solution for xy^'+2y=3cos(x) ?
The solution for xy^'+2y=3cos(x) is y=(3(xsin(x)+cos(x)))/(x^2)+(c_{1})/(x^2)