What is the solution for 11(t+1)y^'-9y=18t ?

The solution for 11(t+1)y^'-9y=18t is y=9t+11+c_{1}(t+1)^{9/11}

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11(t+1)y^'-9y=18t

11 (t + 1) y^{^{'}} - 9 y = 18 t

y = 9 t + 11 + c_{1} (t + 1)^{\frac{9}{11}}

Solution steps

11 (t + 1) y^{'} (t) - 9 y = 18 t

Solve linear ODE:

y = 9 t + 11 + c_{1} (t + 1)^{\frac{9}{11}}

y = 9 t + 11 + c_{1} (t + 1)^{\frac{9}{11}}

y^{^{'}} + \frac{1}{t} y = e^{- t^{2}}, y (1) = 12

\frac{d}{d x} (lo g_{2} ((5 x - 1)^{3}))

\frac{d}{d x} (\frac{x}{x + \frac{b}{x}})

x \to 3 + lim (a^{2} x - 12 a)

(2 x - 1) d x + (5 y + 4) d y = 0

What is the solution for 11(t+1)y^'-9y=18t ?
The solution for 11(t+1)y^'-9y=18t is y=9t+11+c_{1}(t+1)^{9/11}