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

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

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

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

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

Apply initial conditions:

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

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

t \to 0 lim (\frac{1}{2} e^{- 2 t})

\frac{\partial}{\partial x} (4 x + 6 y)

\int e^{- (x + y)} d y

\frac{d}{d x} (\frac{2}{2 - cos ( π x )})

\frac{\partial}{\partial x} (- x ln (x) + x ln (y) + y)

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