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

The solution for y(t+2)y^'+1=-t is y=sqrt(-2t-4+2ln(t+2)+c_{1)},y=-sqrt(-2t-4+2ln(t+2)+c_{1)}

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

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

y = - 2 t - 4 + 2 ln (t + 2) + c_{1}, y = - - 2 t - 4 + 2 ln (t + 2) + c_{1}

Solution steps

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

Solve separable ODE:

y = - 2 t - 4 + 2 ln (t + 2) + c_{1}, y = - - 2 t - 4 + 2 ln (t + 2) + c_{1}

y = - 2 t - 4 + 2 ln (t + 2) + c_{1}, y = - - 2 t - 4 + 2 ln (t + 2) + c_{1}

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\frac{d}{d x} ((x)^{2} (4 x + 1))

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\frac{d}{d x} (\frac{1}{x ^{2}} + \frac{1}{x} + x)

What is the solution for y(t+2)y^'+1=-t ?
The solution for y(t+2)y^'+1=-t is y=sqrt(-2t-4+2ln(t+2)+c_{1)},y=-sqrt(-2t-4+2ln(t+2)+c_{1)}