Matematické Fórum

stuart clark

If $y=f_{1}(x)$ and $y=f_{2}(x)$ are two solution of the equation $ydx+dy=e^{-x}y^2dy$

where $f_{1}(0)=1,f_{2}(x)=-1.$ Then number of solution of $f_{1}(x)f_{2}(x)+x^2=0$

stuart clark

Thanks friends got it.

$ydx+dy = e^{-x}y^2dy\Rightarrow e^xydx+e^xdy=y^2dy$

$\displaystyle d(e^xy)=y^2dy+C\Rightarrow \int d(e^xy) = \int y^2dy$

$\displaystyle \Longrightarrow e^xy=\frac{y^3}{3}+C$

from condition, $\displaystyle e^{x}y=\frac{y^3}{3}\pm \frac{2}{3}$

So $\displaystyle 3e^{x}y_{1}=y^3_{1}+2$ and $\displaystyle 3e^{x}y_{2}=y^3_{2}-2$

So $\displaystyle 9e^{2x}y_{1}y_{2}=y^3_{1}y^3_{2}-4$ and given $y_{1}y_{2}=-x^2$

So $\displaystyle -9x^22e^{2x}=x^6-4\Rightarrow 4-x^6=9x^{2}\cdot e^{2x}$

Now drawing these two graph using derivative test, We have $2$ real solution.

Matematické Fórum

#1 17. 04. 2019 11:52 — Editoval stuart clark (18. 04. 2019 13:00)

no. of real solution

#2 18. 04. 2019 13:01 — Editoval stuart clark (18. 04. 2019 13:02)

Re: no. of real solution

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