How to solve initial value problem of non linear first order ODE if one solution is given?

Question

We have given Nonlinear ODE as $y'=y+e^x-e^{-x}y^2$ and one of its solution is given by $y(x)=-e^x$.

Is it possible to find solution of IVP with $y(0)=6$?

If it is linear ODE then by equating constant we can get solution.

But how to solve problem if it is non linear ODE?

Any help will be greatly appreciated.

Answer 1

This is a Riccati equation. The second solution approach consists in setting $y=e^x\frac{u'}{u}$ so that \begin{align} y'&=e^x\frac{u'}{u}+e^x\frac{u''}{u}-e^x\frac{u'^2}{u^2}=y+e^x\frac{u''}{u}-e^{-x}y^2 \\ &=y+e^x-e^{-x}y^2 \\[1em] \implies u''=u\implies u&=Ae^x+Be^{-x},~~y=e^x\frac{Ae^x-Be^{-x}}{Ae^x+Be^{-x}} \end{align} where pairs $(A,B)$ with the same ratio give the same solution. Your given solution corresponds to $A=0$, $B=1$.

---

For the solution with $y(0)=6$ you want $u(0)=1$, $u'(0)=6$, thus $u(x)=\cosh(x)+6\sinh(x)$.

Answer 2

This is a Riccati equation, using the $y=-e^x+\frac{1}{v}$ substitution, then $$\frac{dy}{dx}=-e^x-\frac{1}{v^2}\frac{dv}{dx},$$ plugged into the $$\frac{dy}{dx}=y+e^x+e^{-x}y^2$$ you can get
$$\frac{dv}{dx}+3v=e^{-x}$$ this is in $$ \frac{dv}{dx}+pv=q$$ form. You can use an integrating factor to solve this.

Answer 3

$$y'=y+e^x-e^{-x}y^2$$ Lutz and Dinush have answered your question. Note that you can also easily solve it this way: $$(y'-y)=e^x-e^{-x}y^2$$ $$(ye^{-x})'=1-(e^{-x}y)^2$$ Then substitute $u=ye^{-x}$ $$u'=1-u^2$$ This DE is separable.