What is the solution for (dy)/(dt)=((te^t))/(ysqrt(5+y^2)) ?

The solution for (dy)/(dt)=((te^t))/(ysqrt(5+y^2)) is

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(dy)/(dt)=((te^t))/(ysqrt(5+y^2))

\frac{d y}{d t} = \frac{( t e ^{t} )}{y 5 + 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} - 5, 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} - 5

Solution steps

\frac{d y}{d t} = \frac{( t e ^{t} )}{y 5 + 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} - 5, 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} - 5

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} - 5, 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} - 5

\frac{\partial}{\partial x} (c x)

f (x) = \frac{8}{x}, at x = 2

\int \frac{7 x ^{2} + 13 x}{x ^{3} + x ^{2} - x - 1} d x

\frac{\partial}{\partial x} (6 x^{\frac{1}{3}} y^{\frac{2}{3}})

\frac{d}{d a} (\frac{cos ^{2} ( a )}{sin ( a ) - 1})

What is the solution for (dy)/(dt)=((te^t))/(ysqrt(5+y^2)) ?
The solution for (dy)/(dt)=((te^t))/(ysqrt(5+y^2)) is