What is the solution for yy^'=e^{2x}+c ?

The solution for yy^'=e^{2x}+c is y=sqrt(e^{2x)+2cx+c_{1}},y=-sqrt(e^{2x)+2cx+c_{1}}

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yy^'=e^{2x}+c

y y^{^{'}} = e^{2 x} + c

y = e^{2 x} + 2 c x + c_{1}, y = - e^{2 x} + 2 c x + c_{1}

Solution steps

y y^{'} (x) = e^{2 x} + c

Solve separable ODE:

y = e^{2 x} + 2 c x + c_{1}, y = - e^{2 x} + 2 c x + c_{1}

y = e^{2 x} + 2 c x + c_{1}, y = - e^{2 x} + 2 c x + c_{1}

y^{^{'}} + y sin (x) = sin (x)

(1 + x) y^{^{'}} - y = 0

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

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

\frac{d p}{d t} = p (2 p - 1)

What is the solution for yy^'=e^{2x}+c ?
The solution for yy^'=e^{2x}+c is y=sqrt(e^{2x)+2cx+c_{1}},y=-sqrt(e^{2x)+2cx+c_{1}}