What is the solution for 1/t (dy)/(dt)-1/(1+t^2)=0 ?

The solution for 1/t (dy)/(dt)-1/(1+t^2)=0 is y= 1/2 ln(t^2+1)+c_{1}

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1/t (dy)/(dt)-1/(1+t^2)=0

\frac{1}{t} \frac{d y}{d t} - \frac{1}{1 + t ^{2}} = 0

y = \frac{1}{2} ln (t^{2} + 1) + c_{1}

Solution steps

\frac{1}{t} \frac{d y}{d t} - \frac{1}{1 + t ^{2}} = 0

Substitute

\frac{d y}{d t}

with

y^{'} (t)

\frac{1}{t} y^{^{'}} (t) - \frac{1}{1 + t ^{2}} = 0

Isolate

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

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y = \int \frac{t}{t ^{2} + 1} d t

\int \frac{t}{t ^{2} + 1} d t = \frac{1}{2} ln (t^{2} + 1) + c_{1}

y = \frac{1}{2} ln (t^{2} + 1) + c_{1}

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What is the solution for 1/t (dy)/(dt)-1/(1+t^2)=0 ?
The solution for 1/t (dy)/(dt)-1/(1+t^2)=0 is y= 1/2 ln(t^2+1)+c_{1}