What is the solution for y^{''}+4y^'+5y= 1/x e^{2x}sin(x) ?

The solution for y^{''}+4y^'+5y= 1/x e^{2x}sin(x) is

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+4y^'+5y= 1/x e^{2x}sin(x)

y^{^{''}} + 4 y^{^{'}} + 5 y = \frac{1}{x} e^{2 x} sin (x)

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) + e^{- 2 x} cos (x) \cdot \int - \frac{sin ^{2} ( x )}{e ^{- 4 x} x} d x + e^{- 2 x} sin (x) \cdot \int \frac{sin ( x ) cos ( x )}{e ^{- 4 x} x} d x

Solution steps

y^{''} (x) + 4 y^{'} (x) + 5 y = \frac{1}{x} e^{2 x} sin (x)

Solve linear ODE:

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) + e^{- 2 x} cos (x) \cdot \int - \frac{sin ^{2} ( x )}{e ^{- 4 x} x} d x + e^{- 2 x} sin (x) \cdot \int \frac{sin ( x ) cos ( x )}{e ^{- 4 x} x} d x

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) + e^{- 2 x} cos (x) \cdot \int - \frac{sin ^{2} ( x )}{e ^{- 4 x} x} d x + e^{- 2 x} sin (x) \cdot \int \frac{sin ( x ) cos ( x )}{e ^{- 4 x} x} d x

y^{^{''}} + 4 y = - 16 e^{- 2 x} + 2 sin (x)

y^{^{''}} + 4 y^{^{'}} = 8 x + 1

y^{^{''}} - y = 2 (e^{x} e^{- x})^{- 1}

4 y^{^{''}} - y = x e^{\frac{1}{2} x}

y^{^{''}} + 2 y^{^{'}} + y = 3 cos (t)

What is the solution for y^{''}+4y^'+5y= 1/x e^{2x}sin(x) ?
The solution for y^{''}+4y^'+5y= 1/x e^{2x}sin(x) is