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

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

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(dy)/(dt)= t/(t^2y+y)

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

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

Solution steps

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

Solve separable ODE:

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

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

\frac{d ^{2}}{d x ^{2}} ((x^{2} + 6)^{4})

\int x^{\frac{- 1}{2}} d x

\int \frac{cos ^{2} ( x )}{sin ^{4} ( x )} d x

n = 3 \sum \infty \frac{5}{9 ^{n}}

\frac{\partial}{\partial y} (x - y^{3} + y^{2} sin (x))

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