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

The solution for y^'=((t^2+ty+y^2))/(t^2) is y=tan(ln(t)+c_{1})t

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y^'=((t^2+ty+y^2))/(t^2)

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

y = tan (ln (t) + c_{1}) t

Solution steps

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

Solve using substitution:

y = tan (ln (t) + c_{1}) t

y = tan (ln (t) + c_{1}) t

h \to 0 lim (48 + 12 h + h^{2})

y = \frac{2 x - 1}{3 x + 5}

n = 1 \sum \infty \frac{6}{( - 5 ) ^{n}}

y^{^{''}} + 256 y = sec (16 x), y (0) = - 5, y^{^{'}} (0) = - 6

\int \frac{cos ( 5 x )}{3 + sin ( 5 x )} d x

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