What is the solution for y^{''}+2y-1(y^')2=y^' ?

The solution for y^{''}+2y-1(y^')2=y^' is y=c_{1}e^{2t}+c_{2}e^t

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y^{''}+2y-1(y^')2=y^'

y^{^{''}} + 2 y - 1 (y^{^{'}}) 2 = y^{^{'}}

y = c_{1} e^{2 t} + c_{2} e^{t}

Solution steps

y^{''} (t) + 2 y - 1 (y^{'} (t)) 2 = y^{'} (t)

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{t}

y = c_{1} e^{2 t} + c_{2} e^{t}

n = 1 \sum \infty 9

x \to 0 - lim (\frac{8 x}{x})

\int sin (\frac{n x}{2}) d x

\frac{d}{d t} (e^{t} - 1)

\int \frac{x ^{4} + x}{x + 1} d x

What is the solution for y^{''}+2y-1(y^')2=y^' ?
The solution for y^{''}+2y-1(y^')2=y^' is y=c_{1}e^{2t}+c_{2}e^t