What is the solution for (dy)/(dt)= t/(t^2y+y),y(0)=-9 ?

The solution for (dy)/(dt)= t/(t^2y+y),y(0)=-9 is y=-sqrt(ln(t^2+1)+81)

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(dy)/(dt)= t/(t^2y+y),y(0)=-9

\frac{d y}{d t} = \frac{t}{t ^{2} y + y}, y (0) = - 9

y = - ln (t^{2} + 1) + 81

Solution steps

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

Solve separable ODE:

y = - ln (t^{2} + 1) + 81

y = - ln (t^{2} + 1) + 81

\frac{\partial}{\partial z} (x^{2} - x yz)

y^{^{''}} - y^{^{'}} - 42 y = 0

\frac{d}{d x} ((3 x + 2)^{x})

y^{^{''}} + y^{^{'}} + 4 y = 0

6 x sin (x)

What is the solution for (dy)/(dt)= t/(t^2y+y),y(0)=-9 ?
The solution for (dy)/(dt)= t/(t^2y+y),y(0)=-9 is y=-sqrt(ln(t^2+1)+81)