What is the solution for (1+y)dy=2tdt ?

The solution for (1+y)dy=2tdt is y=-1+sqrt(2(t^2+c_{1))+1},y=-sqrt(2(t^2+c_{1))+1}-1

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(1+y)dy=2tdt

(1 + y) d y = 2 t d t

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

Solution steps

(1 + y) d y = 2 t d t

Solve separable ODE:

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

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

\int \frac{x ^{2}}{16 - x ^{2}} d x

3 1 + x

(x - 1)^{\frac{1}{2}}

x + y

\int \frac{x}{3 x - 2} d x

What is the solution for (1+y)dy=2tdt ?
The solution for (1+y)dy=2tdt is y=-1+sqrt(2(t^2+c_{1))+1},y=-sqrt(2(t^2+c_{1))+1}-1