How to solve Cauchy problem for $y'+2y= \exp(x)y^2 \quad y(1) = 1$?

Question

The original equation is $$y'+2y= \exp(x)y^2 \quad y(1) = 1$$

Well, this one is Bernulli's one, so I've put $z = \frac{1}{z}$ therefore $z' = -\frac{y'}{y^2}$

and reduced it to the first-order non-linear equation:

$$z'-2z = -e^x$$

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I've applied Lagrange's method

and found the solution (included variabled parameter):

$$y = \left(-\frac{1}{3}e^{-3x} + \alpha\right)e^{2x}$$

But I suspect some mistakes in the solution and not sure how to handle Cauchy's problem here :/

Answer 1

You get $$ (e^{-2x}z(x))'=e^{-2x}(z'(x)-2z(x))=-e^{-x} $$ so that with $z(1)=1$ $$ e^{-2x}z(x)-e^{-2}=e^{-x}-e^{-1} \implies z(x)=e^x+(e^{-2}-e^{-1})e^{2x} \\~\\ \implies y(x)=\frac{e}{e^{x+1}-(e-1)e^{2x-1}}. $$