What is the solution for y^{''}-10y^'+25y=3x+5 ?

The solution for y^{''}-10y^'+25y=3x+5 is y=c_{1}e^{5x}+c_{2}xe^{5x}+(3x)/(25)+31/125

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}-10y^'+25y=3x+5

y^{^{''}} - 10 y^{^{'}} + 25 y = 3 x + 5

y = c_{1} e^{5 x} + c_{2} x e^{5 x} + \frac{3 x}{25} + \frac{31}{125}

Solution steps

y^{''} (x) - 10 y^{'} (x) + 25 y = 3 x + 5

Solve linear ODE:

y = c_{1} e^{5 x} + c_{2} x e^{5 x} + \frac{3 x}{25} + \frac{31}{125}

y = c_{1} e^{5 x} + c_{2} x e^{5 x} + \frac{3 x}{25} + \frac{31}{125}

What is the solution for y^{''}-10y^'+25y=3x+5 ?
The solution for y^{''}-10y^'+25y=3x+5 is y=c_{1}e^{5x}+c_{2}xe^{5x}+(3x)/(25)+31/125