What is the solution for (dy)/(dt)=(te^t)/(ysqrt(4+y^2)) ?

The solution for (dy)/(dt)=(te^t)/(ysqrt(4+y^2)) is

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(dy)/(dt)=(te^t)/(ysqrt(4+y^2))

\frac{d y}{d t} = \frac{t e ^{t}}{y 4 + y ^{2}}

y = 3 18 c_{1} e^{t} t + 9 e^{2 t} t^{2} - 18 e^{2 t} t - 18 c_{1} e^{t} + 9 e^{2 t} + 9 c_{1}^{2} - 4, y = - 3 18 c_{1} e^{t} t + 9 e^{2 t} t^{2} - 18 e^{2 t} t - 18 c_{1} e^{t} + 9 e^{2 t} + 9 c_{1}^{2} - 4

Solution steps

\frac{d y}{d t} = \frac{t e ^{t}}{y 4 + y ^{2}}

Solve separable ODE:

y = 3 18 c_{1} e^{t} t + 9 e^{2 t} t^{2} - 18 e^{2 t} t - 18 c_{1} e^{t} + 9 e^{2 t} + 9 c_{1}^{2} - 4, y = - 3 18 c_{1} e^{t} t + 9 e^{2 t} t^{2} - 18 e^{2 t} t - 18 c_{1} e^{t} + 9 e^{2 t} + 9 c_{1}^{2} - 4

y = 3 18 c_{1} e^{t} t + 9 e^{2 t} t^{2} - 18 e^{2 t} t - 18 c_{1} e^{t} + 9 e^{2 t} + 9 c_{1}^{2} - 4, y = - 3 18 c_{1} e^{t} t + 9 e^{2 t} t^{2} - 18 e^{2 t} t - 18 c_{1} e^{t} + 9 e^{2 t} + 9 c_{1}^{2} - 4

f (x) = \frac{e ^{x} + 1}{e ^{x} - 1}

\int x sin (8 - x) d x

x \to 0 lim (\frac{2 - 2}{2})

\int_{0}^{3} x sin (3 - x) d x

\frac{d}{d x} (2 + x^{2} + x)

What is the solution for (dy)/(dt)=(te^t)/(ysqrt(4+y^2)) ?
The solution for (dy)/(dt)=(te^t)/(ysqrt(4+y^2)) is