What is the solution for 2y^{''}-3y^'=9x^2+10 ?

The solution for 2y^{''}-3y^'=9x^2+10 is y=c_{1}e^{(3x)/2}+c_{2}-x^3-2x^2-6x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2y^{''}-3y^'=9x^2+10

2 y^{^{''}} - 3 y^{^{'}} = 9 x^{2} + 10

y = c_{1} e^{\frac{3 x}{2}} + c_{2} - x^{3} - 2 x^{2} - 6 x

Solution steps

2 y^{''} (x) - 3 y^{'} (x) = 9 x^{2} + 10

Solve linear ODE:

y = c_{1} e^{\frac{3 x}{2}} + c_{2} - x^{3} - 2 x^{2} - 6 x

y = c_{1} e^{\frac{3 x}{2}} + c_{2} - x^{3} - 2 x^{2} - 6 x

y^{^{''}} - y^{^{'}} - 2 y = sin (t)

x^{^{''}} + 14 x^{^{'}} + 98 x = 1960

y^{^{''}} + 4 y = sin (x), y (0) = 1, y^{^{'}} (0) = 2

x^{^{''}} + 4 x = 289 + e^{t} \cdot sin (2 t)

y^{^{''}} - 10 y^{^{'}} + 125 y = - 4 sin (3 t)

What is the solution for 2y^{''}-3y^'=9x^2+10 ?
The solution for 2y^{''}-3y^'=9x^2+10 is y=c_{1}e^{(3x)/2}+c_{2}-x^3-2x^2-6x