What is the solution for y^{''}-5y^'+6y=10sin(2x)-50cos(2x) ?

The solution for y^{''}-5y^'+6y=10sin(2x)-50cos(2x) is y=c_{1}e^{3x}+c_{2}e^{2x}+5sin(2x)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-5y^'+6y=10sin(2x)-50cos(2x)

y^{^{''}} - 5 y^{^{'}} + 6 y = 10 sin (2 x) - 50 cos (2 x)

y = c_{1} e^{3 x} + c_{2} e^{2 x} + 5 sin (2 x)

Solution steps

y^{''} (x) - 5 y^{'} (x) + 6 y = 10 sin (2 x) - 50 cos (2 x)

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{2 x} + 5 sin (2 x)

y = c_{1} e^{3 x} + c_{2} e^{2 x} + 5 sin (2 x)

y^{^{''}} - 10 y^{^{'}} + 25 y = t^{- 4} e^{5 t}

y^{^{''}} - 2 y^{^{'}} + 17 y = 289 x^{2} + 9

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

- x^{^{''}} - 4 x = t^{2}

y^{^{''}} + 3 y^{^{'}} + 2 y = cos (x)

What is the solution for y^{''}-5y^'+6y=10sin(2x)-50cos(2x) ?
The solution for y^{''}-5y^'+6y=10sin(2x)-50cos(2x) is y=c_{1}e^{3x}+c_{2}e^{2x}+5sin(2x)