What is the solution for t^2y^'+y^2=ty,y(1)=-1/2 ?

The solution for t^2y^'+y^2=ty,y(1)=-1/2 is y= t/(ln(t)-2)

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t^2y^'+y^2=ty,y(1)=-1/2

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

y = \frac{t}{ln ( t ) - 2}

Solution steps

t^{2} y^{'} (t) + y^{2} = t y

Solve Bernoulli ODE:

y = \frac{t}{ln ( t ) - 2}

y = \frac{t}{ln ( t ) - 2}

\frac{\partial}{\partial x} (ln (x - y^{2}))

\frac{d y}{d x} = \frac{4 sec ( y )}{( x - 6 ) ^{2}}

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

\frac{d x}{d y} = 4 (x^{2} + 1)

x \to \infty lim (\frac{4 x ^{3} + 8}{4 x ^{3} + 64 x ^{6} + 3})

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