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Hi I need to solve $$u_x +xyu_y +(2x^2z\ln|y|)u_z = 0$$

Μy try: I wrote the equations: $$ dx = \frac{dy}{xy} = \frac{dz}{2x^2z\ln|y|} $$ so the first surface is $\phi_1 = u$

Second surface: $$ dx\cdot x = \frac{dy}{y} \Rightarrow 0.5x^2 + C_1= \ln|y| \Rightarrow C = y\cdot e^{-0.5x^2}$$ $$ \phi_2 = ye^{\frac{x^2}{2}}$$

Third surface (I think here lies the problem): $$ 2x^2\ln|y| = \frac{dz}{z} \Rightarrow \ln|y|(\frac{2x^3}{3} + C) = \ln|z| \Rightarrow \\ \Rightarrow y^{\frac{2x^3}{3}}\cdot y^C = z $$ Now I substitue y: $$ z \cdot y^{\frac{-2x^3}{3}} = Ce^{0.5x^2} \Rightarrow z \cdot y^{\frac{-2x^3}{3}}e^{-0.5x^2} = C $$ So the third surface is: $$ \phi_3 = zy^{\frac{-2x^3}{3}}e^{-0.5x^2} $$ Which is a mistake... according to the solution: $\phi_3 = zy^{\frac{-2x^3}{3}}e^{\frac{2x^5}{15}}$

because the solution is $$u(x, y) = f(zy^{\frac{-2x^3}{3}}e^{\frac{2x^5}{15}}, ye^{\frac{x^2}{2}})$$

I can't figure out where I went wrong, any help/hints would be appreciated!

Giorgos Giapitzakis
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Edward Josef
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1 Answers1

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$$ dx = \frac{dy}{xy} = \frac{dz}{2x^2z\ln|y|} \quad\text{is OK.}$$

I agree with $ \quad 0.5x^2 + C_1= \ln|y| \Rightarrow C = y\cdot e^{-0.5x^2}$ $$ \phi_2 = ye^{-\frac{x^2}{2}}$$ $\phi_2 = ye^{\frac{x^2}{2}}$ is not correct.

For the third surface, you forgot $dx$ in $ 2x^2\ln|y| = \frac{dz}{z} $ which should be $$ 2x^2\ln|y|dx = \frac{dz}{z}$$ with $\ln|y|=0.5x^2 + C_1$ $$ 2x^2(0.5x^2 + C_1)dx = \frac{dz}{z}$$ $$ (x^4 + 2C_1x^2)dx = \frac{dz}{z}$$ Introducing another constant $C\neq C_1$ is a cause of confusion.

I suppose that you can take it from here.

Note : There is a typo in the expected solution $u(x, y) = f(zy^{\frac{-2x^3}{3}}e^{\frac{2x^5}{15}}, ye^{\frac{x^2}{2}})$ which should be : $$u(x, y) = f(zy^{\frac{-2x^3}{3}}e^{\frac{2x^5}{15}}, ye^{-\frac{x^2}{2}})$$

JJacquelin
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  • Hey, thank you! just a quick follow up, I get eventually that $\phi_3 = z\ e^{\frac{2x^5}{15}} \ y^{\frac{-2x^3}{3}} $ which is fine but the constant on the other side of it is $0$ is that a problem or is it fine ?, thanks again ! – Edward Josef Oct 13 '21 at 15:05
  • What do mean by "The constant on the other side" ? – JJacquelin Oct 13 '21 at 15:57