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

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

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

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

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

Solution steps

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

Solve separable ODE:

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

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

\int x ln (2 x - 3) d x

\int \frac{1}{x ^{2} + 10 x + 25} d x

n \to \infty lim (\frac{1}{n !})

x \to 0 lim (\frac{1}{x ^{2} ( x + 8 )})

y^{^{''}} + 4 y = sin (2 x)

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