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I'm new to solving differential equations.

How would we go about solving a differential equation like

$y'=-4+5y-y^2$ ?

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  • $\begingroup$ It’s separable: $y’=f(y)$ $\endgroup$ Jan 24, 2021 at 1:27

2 Answers 2

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$$\frac{dy}{dt} = -4+5y-y^2$$

$$\frac{dy}{-4+5y-y^2}=dt$$

$$\int \frac{dy}{4+5y-y^2} = \int dt$$

$$\frac{1}{3}\text{ln}(|\frac{y-1}{y-4}|)=t+c_1$$

$$\text{ln}(|\frac{3}{y-4} + 1|)=3t+c_2$$

From here just you can isolate for y, or just leave it implicit.

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One more way, use $y=u'/u$, then $$y'=-4+5y-y^2 \implies \frac{u u''-u'^2}{u^2}=-4+5\frac{u'}{u}-\frac{u'^2}{u^2} $$ $$\implies u''-5u'+4=0$$ Let $u=e^{mx}$, then $m^2-5m+2=0 \implies m=2,3$. So $u(x)=C_1e^{2x}+C_2 e^{3x}$, which gives $$y(x)=\frac{2C_1 e^{2x}+ 3 C_2 e^{3x}}{C_1 e^{2x}+ C_2 e^{3x}} =\frac{2+3Ce^{x}}{1+Ce^{x}}$$

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