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

The solution for y^{''}+4y=g(x),y(0)=1,y^'(0)=3 is y=cos(2x)+(sin(2x)(12-g))/8+(gx)/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+4y=g(x),y(0)=1,y^'(0)=3

y^{^{''}} + 4 y = g (x), y (0) = 1, y^{^{'}} (0) = 3

y = cos (2 x) + \frac{sin ( 2 x ) ( 12 - g )}{8} + \frac{gx}{4}

Solution steps

y^{''} (x) + 4 y = g (x)

Solve linear ODE:

y = cos (2 x) + \frac{sin ( 2 x ) ( 12 - g )}{8} + \frac{gx}{4}

y = cos (2 x) + \frac{sin ( 2 x ) ( 12 - g )}{8} + \frac{gx}{4}

y^{^{''}} + 4 y^{^{'}} - 5 y = 10 x^{2} + 9 e^{x}

(d - 1)^{2} (d - 3)^{3} y = e^{3} x

y^{^{''}} - 2 y^{^{'}} - 3 y = 15 t e^{2 t}

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

y^{^{''}} + y^{^{'}} - 6 y = 50 x e^{x}

What is the solution for y^{''}+4y=g(x),y(0)=1,y^'(0)=3 ?
The solution for y^{''}+4y=g(x),y(0)=1,y^'(0)=3 is y=cos(2x)+(sin(2x)(12-g))/8+(gx)/4