What is the solution for (dy)/(dt)=e^{6y+4t},y(0)=14 ?

The solution for (dy)/(dt)=e^{6y+4t},y(0)=14 is y=-(ln(-3/2 e^{4t}+1/(e^{84)}+3/2))/6

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(dy)/(dt)=e^{6y+4t},y(0)=14

\frac{d y}{d t} = e^{6 y + 4 t}, y (0) = 14

y = - \frac{ln ( - \frac{3}{2} e ^{4 t} + \frac{1}{e ^{84}} + \frac{3}{2} )}{6}

Solution steps

\frac{d y}{d t} = e^{6 y + 4 t}

Solve separable ODE:

y = - \frac{ln ( - \frac{3}{2} e ^{4 t} + \frac{1}{e ^{84}} + \frac{3}{2} )}{6}

y = - \frac{ln ( - \frac{3}{2} e ^{4 t} + \frac{1}{e ^{84}} + \frac{3}{2} )}{6}

y^{^{'}} = x^{2} y^{2} - y^{2} + x^{2} - 1

x^{2} \frac{d y}{d x} = 5 y

\frac{d x}{d t} = 14 - 0.84 x

2 x + 4 y \frac{d y}{d x} - c = 0

\frac{d y}{d x} = (y + 5) (2 x + 1)

What is the solution for (dy)/(dt)=e^{6y+4t},y(0)=14 ?
The solution for (dy)/(dt)=e^{6y+4t},y(0)=14 is y=-(ln(-3/2 e^{4t}+1/(e^{84)}+3/2))/6