What is the solution for ty^'+(2t^2+1)y=3t ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

ty^'+(2t^2+1)y=3t

t y^{^{'}} + (2 t^{2} + 1) y = 3 t

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

Solution steps

t y^{'} (t) + (2 t^{2} + 1) y = 3 t

Solve linear ODE:

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

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

x, - 1, 2

\int \frac{1}{z ^{3} z ^{2} - 36} d z

f (x) = e^{4}

\int \frac{1 + cos ( 4 x )}{2} d x

\int 13 y cos (3 x) + 3 sin (y) d x

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