What is the solution for ty^'+(t+1)y=t,y(ln(6))=1 ?

The solution for ty^'+(t+1)y=t,y(ln(6))=1 is y=1-1/t+6/(e^tt)

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ty^'+(t+1)y=t,y(ln(6))=1

t y^{^{'}} + (t + 1) y = t, y (ln (6)) = 1

y = 1 - \frac{1}{t} + \frac{6}{e ^{t} t}

Solution steps

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

Solve linear ODE:

y = 1 - \frac{1}{t} + \frac{6}{e ^{t} t}

y = 1 - \frac{1}{t} + \frac{6}{e ^{t} t}

\frac{\partial}{\partial x} (3 x^{2} + 7 y^{2})

(2.3) (4.7)

\int - ln ∣ 1 - x ∣ d x

\int_{0}^{H} z^{2} (1 - z)^{2} d z

(x^{2} + y) + (x + e^{y}) y^{^{'}} = 0

What is the solution for ty^'+(t+1)y=t,y(ln(6))=1 ?
The solution for ty^'+(t+1)y=t,y(ln(6))=1 is y=1-1/t+6/(e^tt)