What is the solution for y^'+3ty=t,y(0)=1 ?

The solution for y^'+3ty=t,y(0)=1 is y=(2e^{-(3t^2)/2})/(3)+1/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'+3ty=t,y(0)=1

y^{^{'}} + 3 t y = t, y (0) = 1

y = \frac{2 e ^{- \frac{3 t ^{2}}{2}}}{3} + \frac{1}{3}

Solution steps

y^{'} (t) + 3 t y = t

Solve separable ODE:

y = \frac{2 e ^{- \frac{3 t ^{2}}{2}}}{3} + \frac{1}{3}

y = \frac{2 e ^{- \frac{3 t ^{2}}{2}}}{3} + \frac{1}{3}

\int \frac{9}{x ( ln ( x ) ) ^{2}} d x

x \to 0 lim (x (cot (2 x)))

x, x^{2}

\frac{d}{d x} (1 + x^{3})

f (x) = 5 x + 6, at x = 2

What is the solution for y^'+3ty=t,y(0)=1 ?
The solution for y^'+3ty=t,y(0)=1 is y=(2e^{-(3t^2)/2})/(3)+1/3