What is the solution for ty^{''}+4y^'=t^2 ?

The solution for ty^{''}+4y^'=t^2 is y=c_{1}+(c_{2})/(t^3)+(t^3)/(18)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

ty^{''}+4y^'=t^2

t y^{^{''}} + 4 y^{^{'}} = t^{2}

y = c_{1} + \frac{c _{2}}{t ^{3}} + \frac{t ^{3}}{18}

Solution steps

t y^{''} (t) + 4 y^{'} (t) = t^{2}

Solve linear ODE:

y = c_{1} + \frac{c _{2}}{t ^{3}} + \frac{t ^{3}}{18}

y = c_{1} + \frac{c _{2}}{t ^{3}} + \frac{t ^{3}}{18}

What is the solution for ty^{''}+4y^'=t^2 ?
The solution for ty^{''}+4y^'=t^2 is y=c_{1}+(c_{2})/(t^3)+(t^3)/(18)