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

The solution for ty^'+(t+1)y=t,y(ln(2))=1 is y=1-1/t+2/(e^tt)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

ty^'+(t+1)y=t,y(ln(2))=1

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

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

Solution steps

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

Solve linear ODE:

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

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

\int (6 sin (2 x) + sec^{2} (x)) d x

\frac{d ^{2}}{d x ^{2}} (\frac{x ^{2}}{4 + x})

\int (125 x^{9}) d x

\frac{d}{d x} (2 cos (x^{2}) sin (2 x))

What is the solution for ty^'+(t+1)y=t,y(ln(2))=1 ?
The solution for ty^'+(t+1)y=t,y(ln(2))=1 is y=1-1/t+2/(e^tt)