What is the solution for xy^'=y+2x^2sin(x),y(pi)=0 ?

The solution for xy^'=y+2x^2sin(x),y(pi)=0 is y=-2xcos(x)-2x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

xy^'=y+2x^2sin(x),y(pi)=0

x y^{^{'}} = y + 2 x^{2} sin (x), y (π) = 0

y = - 2 x cos (x) - 2 x

Solution steps

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

Solve linear ODE:

y = - 2 x cos (x) - 2 x

y = - 2 x cos (x) - 2 x

n = 1 \sum \infty \frac{e ^{n}}{n ^{2}}

\frac{\partial}{\partial x} (2 x^{3} + 3 y^{3} - 12 x y)

\int \frac{1}{( x ^{2} - 6 x ) ^{\frac{3}{2}}} d x

\frac{\partial}{\partial x} (x y E (x, y) (x, y)^{\frac{x}{y}})

\int - 2 cos^{2} (x) d x

What is the solution for xy^'=y+2x^2sin(x),y(pi)=0 ?
The solution for xy^'=y+2x^2sin(x),y(pi)=0 is y=-2xcos(x)-2x