<s>[INST] Solve the differential equation : $0.5 \frac{dy}{dx}=4.9-0.1y^2$ 
The question is to solve the differential equation :
$$0.5 \frac{dy}{dx}=4.9-0.1y^2$$


What I have attempted:
$$0.5 \frac{dy}{dx}=4.9-0.1y^2$$
$$ \frac{dy}{dx} = \frac{4.9-0.1y^2}{0.5} $$
$$ \frac{dy}{dx} = 9.8 - 0.2y^2 $$
$$ \int \frac{1}{9.8 - 0.2y^2} dy = \int 1 dx $$
$$ \int \frac{10}{98 - 2y^2} dy = \int 1 dx $$
$$ \int \frac{5}{49 - y^2} dy = \int 1 dx $$
$$ \int \frac{1}{49 - y^2} dy = \int \frac{1}{5} dx $$
How should I continue?
EDIT Thanks @andrenicolas for the hint
By using partial fractions
$$ \int \frac{1}{(7-y)(7+y)} dy = \int \frac{1}{5} dx $$
$$ \frac{1}{14}\int \frac{1}{7-y} + \frac {1}{7+y} dy = \int \frac{1}{5} dx $$
$$ \frac{1}{14} [ ln |y+7| - ln|y-7| ] =  \frac{x}{5} + c$$
$$ [ ln |y+7| - ln|y-7| ] = \frac{14x}{5} + 14c $$
$$ ln\frac{y+7}{y-7} = \frac{14x}{5} + 14c $$
$$ \frac{y+7}{y-7} = Ae^{\frac{14x}{5}} $$ (Letting $A = 14c$)
$$ y + 7 = yAe^{\frac{14x}{5}} - 7Ae^{\frac{14x}{5}} $$
$$ 7Ae^{\frac{14x}{5}} + 7 = yAe^{\frac{14x}{5}} - y $$
$$ 7Ae^{\frac{14x}{5}} + 7 = y(Ae^{\frac{14x}{5}} - 1) $$
$$ y = \frac{7Ae^{\frac{14x}{5}}}{(Ae^{\frac{14x}{5}} - 1)} $$
 [/INST] Continuing your calculations
$$ \int \frac{1}{49 - y^2} dy = \int \frac{1}{5} dx $$
$$  \int \frac{1}{7^2 - y^2} dy = \int \frac{1}{5} dx $$
$$  \int \frac{1}{7 - y} + \frac{1}{7 + y} dy = \int \frac{14}{5} dx $$
$$ \log(\frac{7+y}{7-y}) = \frac{14}{5} x + C$$
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