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

The solution for ty^'+(t+1)y=t,y(ln(3))=1 is y=1-1/t+3/(e^tt)

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

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

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

Solution steps

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

Solve linear ODE:

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

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

x \to 0 lim (x^{4} (cos (\frac{2}{x})))

\int \frac{4}{5} d x

f (x) = 4^{\frac{c}{x}}

\frac{d}{d x} (\frac{1}{4} sin^{4} (x))

x \to \infty lim (1 + 2^{x})

What is the solution for ty^'+(t+1)y=t,y(ln(3))=1 ?
The solution for ty^'+(t+1)y=t,y(ln(3))=1 is y=1-1/t+3/(e^tt)