What is the solution for ty^'+(t+1)y=t,y(ln(5))=1 ?

The solution for ty^'+(t+1)y=t,y(ln(5))=1 is y=1-1/t+5/(e^tt)

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ty^'+(t+1)y=t,y(ln(5))=1

t y^{^{'}} + (t + 1) y = t, y (ln (5)) = 1

y = 1 - \frac{1}{t} + \frac{5}{e ^{t} t}

Solution steps

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

Solve linear ODE:

y = 1 - \frac{1}{t} + \frac{5}{e ^{t} t}

y = 1 - \frac{1}{t} + \frac{5}{e ^{t} t}

\frac{e ^{3 x^{2}}}{5 x}

\frac{d}{d x} (6 (1 - \frac{x}{12})^{2})

y^{2}

t \to 0 lim (\frac{4 t + 2}{t + 1})

(19.3) (20.3)

What is the solution for ty^'+(t+1)y=t,y(ln(5))=1 ?
The solution for ty^'+(t+1)y=t,y(ln(5))=1 is y=1-1/t+5/(e^tt)