What is the solution for y^{''}-5y^'=2x+1 ?

The solution for y^{''}-5y^'=2x+1 is y=c_{1}e^{5x}+c_{2}-(x^2)/5-(7x)/(25)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-5y^'=2x+1

y^{^{''}} - 5 y^{^{'}} = 2 x + 1

y = c_{1} e^{5 x} + c_{2} - \frac{x ^{2}}{5} - \frac{7 x}{25}

Solution steps

y^{''} (x) - 5 y^{'} (x) = 2 x + 1

Solve linear ODE:

y = c_{1} e^{5 x} + c_{2} - \frac{x ^{2}}{5} - \frac{7 x}{25}

y = c_{1} e^{5 x} + c_{2} - \frac{x ^{2}}{5} - \frac{7 x}{25}

y^{^{'''}} - 11 y^{^{''}} + 28 y^{^{'}} = 90 e^{x}

y^{^{''}} + y^{^{'}} = arctan (x)

y^{^{''}} - 2 y^{^{'}} - 8 y = 5 cos (t)

y^{^{''}} + 2 y^{^{'}} - 8 y = 8 x^{2} - 3

\frac{d ^{2} y}{d x ^{2}} + y = - 16 \cdot cos (3 x)

What is the solution for y^{''}-5y^'=2x+1 ?
The solution for y^{''}-5y^'=2x+1 is y=c_{1}e^{5x}+c_{2}-(x^2)/5-(7x)/(25)