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

The solution for (t+2)y^'+y=(1/(t-1)) is y=(ln(t-1))/(t+2)+(c_{1})/(t+2)

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(t+2)y^'+y=(1/(t-1))

(t + 2) y^{^{'}} + y = (\frac{1}{t - 1})

y = \frac{ln ( t - 1 )}{t + 2} + \frac{c _{1}}{t + 2}

Solution steps

(t + 2) y^{'} (t) + y = (\frac{1}{t - 1})

Solve linear ODE:

y = \frac{ln ( t - 1 )}{t + 2} + \frac{c _{1}}{t + 2}

y = \frac{ln ( t - 1 )}{t + 2} + \frac{c _{1}}{t + 2}

x \to 5 - lim (5 - x)

\frac{\partial}{\partial z} (4 x^{3} + z^{2})

7 x^{4}

x, x, x = 0, x = 2

\frac{\partial}{\partial x} (cos^{2} (x))

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