What is the solution for 9t((dy)/(dt))+y=t^2 ?

The solution for 9t((dy)/(dt))+y=t^2 is y=(t^2)/(19)+(c_{1})/(\sqrt[9]{t)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

9t((dy)/(dt))+y=t^2

9 t (\frac{d y}{d t}) + y = t^{2}

y = \frac{t ^{2}}{19} + \frac{c _{1}}{9 t}

Solution steps

9 t (\frac{d y}{d t}) + y = t^{2}

Solve linear ODE:

y = \frac{t ^{2}}{19} + \frac{c _{1}}{9 t}

y = \frac{t ^{2}}{19} + \frac{c _{1}}{9 t}

\frac{\partial}{\partial y} (\frac{e ^{2 y}}{2 x})

\int_{1}^{e} 4 x^{2} ln (x) d x

\frac{d}{d x} (\frac{1}{1 + x})

\frac{d}{d x} (e^{\frac{1}{2}})

\int \frac{1 - x}{x ^{3}} d x

What is the solution for 9t((dy)/(dt))+y=t^2 ?
The solution for 9t((dy)/(dt))+y=t^2 is y=(t^2)/(19)+(c_{1})/(\sqrt[9]{t)}