What is the solution for y^{''}+16y=32+45e^x ?

The solution for y^{''}+16y=32+45e^x is y=c_{1}cos(4x)+c_{2}sin(4x)+2+45/17 e^x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+16y=32+45e^x

y^{^{''}} + 16 y = 32 + 45 e^{x}

y = c_{1} cos (4 x) + c_{2} sin (4 x) + 2 + \frac{45}{17} e^{x}

Solution steps

y^{''} (x) + 16 y = 32 + 45 e^{x}

Solve linear ODE:

y = c_{1} cos (4 x) + c_{2} sin (4 x) + 2 + \frac{45}{17} e^{x}

y = c_{1} cos (4 x) + c_{2} sin (4 x) + 2 + \frac{45}{17} e^{x}

d (x) = x^{2} - 8 x + 15

y^{^{''}} + 3 y^{^{'}} - 4 y = 36 e^{5 t}

y^{^{'''}} - 5 y^{^{''}} + 6 y^{^{'}} = 3 + 7 x

y^{^{''}} - y = \frac{e ^{2 x} - e ^{- 2 x}}{2}

y^{^{''}} + y = 1 - 1 - H (\frac{π}{2} \cdot t)

What is the solution for y^{''}+16y=32+45e^x ?
The solution for y^{''}+16y=32+45e^x is y=c_{1}cos(4x)+c_{2}sin(4x)+2+45/17 e^x