What is the solution for y^{''}+4y=6e^{2x},y(0)=1,y^'(0)=3 ?

The solution for y^{''}+4y=6e^{2x},y(0)=1,y^'(0)=3 is y= 1/4 cos(2x)+3/4 sin(2x)+3/4 e^{2x}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}+4y=6e^{2x},y(0)=1,y^'(0)=3

y^{^{''}} + 4 y = 6 e^{2 x}, y (0) = 1, y^{^{'}} (0) = 3

y = \frac{1}{4} cos (2 x) + \frac{3}{4} sin (2 x) + \frac{3}{4} e^{2 x}

Solution steps

y^{''} (x) + 4 y = 6 e^{2 x}

Solve linear ODE:

y = \frac{1}{4} cos (2 x) + \frac{3}{4} sin (2 x) + \frac{3}{4} e^{2 x}

y = \frac{1}{4} cos (2 x) + \frac{3}{4} sin (2 x) + \frac{3}{4} e^{2 x}

What is the solution for y^{''}+4y=6e^{2x},y(0)=1,y^'(0)=3 ?
The solution for y^{''}+4y=6e^{2x},y(0)=1,y^'(0)=3 is y= 1/4 cos(2x)+3/4 sin(2x)+3/4 e^{2x}