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

The solution for y^'= 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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y^'= t/(t^2y+y)

y^{^{'}} = \frac{t}{t ^{2} y + y}

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

Solution steps

y^{'} (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}

n = 1 \sum \infty n e^{- \frac{n}{8}}

y^{^{''}} + 20 y^{^{'}} = 0, y (0) = - 5, y^{^{'}} (0) = 20

\frac{d ^{2}}{d x ^{2}} (x^{\frac{1}{11}})

\frac{\partial}{\partial x} (x e^{x^{2} + y^{2}})

\frac{\partial}{\partial x} (\frac{e ^{v}}{u ( v , x ) + v ^{2}})

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