What is the solution for y^{''}-10y^'+25y=132x^2(e^{(5x)}) ?

The solution for y^{''}-10y^'+25y=132x^2(e^{(5x)}) is y=c_{1}e^{5x}+c_{2}xe^{5x}+11e^{5x}x^4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-10y^'+25y=132x^2(e^{(5x)})

y^{^{''}} - 10 y^{^{'}} + 25 y = 132 x^{2} (e^{(5 x)})

y = c_{1} e^{5 x} + c_{2} x e^{5 x} + 11 e^{5 x} x^{4}

Solution steps

y^{''} (x) - 10 y^{'} (x) + 25 y = 132 x^{2} (e^{(5 x)})

Solve linear ODE:

y = c_{1} e^{5 x} + c_{2} x e^{5 x} + 11 e^{5 x} x^{4}

y = c_{1} e^{5 x} + c_{2} x e^{5 x} + 11 e^{5 x} x^{4}

y^{^{''}} - 2 y = e^{2 x} + x^{2} - 1

y^{^{''}} - 3 y^{^{'}} = 2 e^{2 x} sin (x)

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

x^{^{''}} + 4 x = 32 sin (4 t)

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

What is the solution for y^{''}-10y^'+25y=132x^2(e^{(5x)}) ?
The solution for y^{''}-10y^'+25y=132x^2(e^{(5x)}) is y=c_{1}e^{5x}+c_{2}xe^{5x}+11e^{5x}x^4