Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


Find the general solution of the equation

$$y\dfrac{\partial z}{\partial x}+2z\dfrac{\partial z}{\partial y}=\frac{y}{x}$$

Then solve the Cauchy problem with Cauchy data

$$x=y^2, \ \ z=2$$

That is, find the integral surface of this equation passing through that curve.

Attempt at solution:

Characteristic system:

$$\frac{dx}{y}=\frac{dy}{2z}=\frac{x\ dz}{y}$$

Integrating $\frac{dx}{y}=\frac{dy}{2z}$ gives


Integrating $\frac{dx}{y}=\frac{x \ dz}{y}$ gives

$$u_2(x,y,z)=C2=\ln x-z$$

Inserting Cauchy data gives:


$$C_2=\ln x-\ln y^2-z+2$$

But I am unsure of what to do next to get the general solution.

share|cite|improve this question

$y\dfrac{\partial z}{\partial x}+2z\dfrac{\partial z}{\partial y}=\dfrac{y}{x}$

$x\dfrac{\partial z}{\partial x}+\dfrac{2xz}{y}\dfrac{\partial z}{\partial y}=1$

Follow the method in

$\dfrac{dz}{dt}=1$ , letting $z(0)=0$ , we have $z=t$

$\dfrac{dx}{dt}=x$ , letting $x(0)=x_0$ , we have $x=x_0e^t$

$\dfrac{dy}{dt}=\dfrac{2xz}{y}=\dfrac{2x_0te^t}{y}$ , we have $y^2=4x_0(t-1)e^t+f(x_0)=4x(z-1)+f(xe^{-z})$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.