What is the solution for x^2y^'+2xy=25cos^2(x) ?

The solution for x^2y^'+2xy=25cos^2(x) is y=(25(2x+sin(2x)))/(4x^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 >

x^2y^'+2xy=25cos^2(x)

x^{2} y^{^{'}} + 2 x y = 25 cos^{2} (x)

y = \frac{25 ( 2 x + sin ( 2 x ) )}{4 x ^{2}} + \frac{c _{1}}{x ^{2}}

Solution steps

x^{2} y^{'} (x) + 2 x y = 25 cos^{2} (x)

Solve linear ODE:

y = \frac{25 ( 2 x + sin ( 2 x ) )}{4 x ^{2}} + \frac{c _{1}}{x ^{2}}

y = \frac{25 ( 2 x + sin ( 2 x ) )}{4 x ^{2}} + \frac{c _{1}}{x ^{2}}

\frac{- 8 x}{x ^{2} + 1}

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

\int \frac{6 x + 7}{( x + 2 ) ^{2}} d x

\int \frac{1}{x ( ln ( x ) ) ^{3}} d x

y^{^{'}} = y (y + 1) x

What is the solution for x^2y^'+2xy=25cos^2(x) ?
The solution for x^2y^'+2xy=25cos^2(x) is y=(25(2x+sin(2x)))/(4x^2)+(c_{1})/(x^2)