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Let $$U_{xy}+U_y=e^{-x}$$

I followed the substitution mentioned here.

Let $V_x+V=e^{-x}$. So now we have $(e^{x}V)_x=e^{-x}$. Integrating w.r.t $x$ we get $$V=-e^{-2x}+e^{-x}c_1(y).$$ Then integrating w.r.t $y$ we get $$U=-ye^{-2x}+ye^{-x}c_1(y)+c_2(x).$$

Is this the correct procedure?

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emka
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  • Why do you ask? You can partial differentiate and check whether you get what you want. Something is probably wrong as differentiating partially your $;U;$ I get $;U_{xy}+U_y=e^{-2x};$ – Timbuc Sep 23 '14 at 14:01

2 Answers2

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$V_x+ V = {\rm e}^{-x}$ implies $({\rm e}^{x}V)_x=1$.

ir7
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$$U_{xy}+U_y=e^{-x}$$ Interation relatively to $y$ gives : $U_x+U=e^{-x}y+f(x)$ any derivable function $f(x)$

Let $U(x,y)=e^{-x}F(x,y)$ $$U_x+U=e^{-x}F_x=e^{-x}y+f(x)$$ $$F_x=y+e^xf(x)=y+g(x)$$ Integration relatively to $x$ gives : $F=xy+G(x)+h(y)$ any derivable functions $G$ and $h$ $$U(x,y)=e^{-x}\left(xy+G(x)+h(y)\right)$$ or $$U(x,y)=e^{-x}\left(xy+h(y)\right)+H(x)$$ where $h$ and $H$ are any derivable functions.

JJacquelin
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