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I tried to solve the PDE $$(y^3x-2x^4)u_x+(2y^4-x^3y)u_y=9u(x^3-y^3).$$ I used the method of characterstics and got the system of ODEs $$x'(t)=(y(t))^3x(t)-2(x(t))^4, \\ y'(t)=2(y(t))^4-2(x(t))^3y(t),\\ u'(t)=9u(t)((x(t))^3-(y(t))^3)$$ but I have no idea how to get any first integrals.

Sz_Z
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1 Answers1

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The system can be written in non-parametric form as: $$\frac{d x}{y^3x-2x^4}=\frac{dy}{2y^4-x^3y}=\frac{du}{9u(x^3-y^3)}=\dfrac{\frac{dx}{x}+\frac{dy}{y}+\frac{du}{3u}}{0}$$ From the last ratio, we have

$$\frac{dx}{x}+\frac{dy}{y}+\frac{du}{3u}=0$$ Integration yields, $$3 x y u=\text{constant}.$$

Arun
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