What is the solution for ydy+(1+y^2)sin(t)dt=0 ?

The solution for ydy+(1+y^2)sin(t)dt=0 is y=sqrt(e^{2cos(t)+c_{1)}-1},y=-sqrt(e^{2cos(t)+c_{1)}-1}

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ydy+(1+y^2)sin(t)dt=0

y d y + (1 + y^{2}) sin (t) d t = 0

y = e^{2 c o s (t) + c_{1}} - 1, y = - e^{2 c o s (t) + c_{1}} - 1

Solution steps

y d y + (1 + y^{2}) sin (t) d t = 0

Solve separable ODE:

y = e^{2 c o s (t) + c_{1}} - 1, y = - e^{2 c o s (t) + c_{1}} - 1

y = e^{2 c o s (t) + c_{1}} - 1, y = - e^{2 c o s (t) + c_{1}} - 1

What is the solution for ydy+(1+y^2)sin(t)dt=0 ?
The solution for ydy+(1+y^2)sin(t)dt=0 is y=sqrt(e^{2cos(t)+c_{1)}-1},y=-sqrt(e^{2cos(t)+c_{1)}-1}