What is the solution for xy^'+2y=e^{(2x^2)} ?

The solution for xy^'+2y=e^{(2x^2)} is y=(e^{2x^2})/(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 >

xy^'+2y=e^{(2x^2)}

x y^{^{'}} + 2 y = e^{(2 x^{2})}

y = \frac{e ^{2 x^{2}}}{4 x ^{2}} + \frac{c _{1}}{x ^{2}}

Solution steps

x y^{'} (x) + 2 y = e^{(2 x^{2})}

Solve linear ODE:

y = \frac{e ^{2 x^{2}}}{4 x ^{2}} + \frac{c _{1}}{x ^{2}}

y = \frac{e ^{2 x^{2}}}{4 x ^{2}} + \frac{c _{1}}{x ^{2}}

\frac{\partial}{\partial x} (x \cdot arcsin (\frac{y}{x}))

\frac{10}{( s + 4 ) ( s + 2 ) ^{3}}

\int cos^{2} (2 x + 1) d x

\frac{\partial}{\partial x} (- 2 tan (x) sin (2 y))

\frac{d}{d x} (- A e^{- x})

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