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Question:

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

$$u_1(x,y,z)=C1=x-\frac{y^2}{4z}$$

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

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

Inserting Cauchy data gives:

$$C_1=y^2-\frac{y^2}{8}-\frac{7}{8}y^2$$

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

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

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1 Answer 1

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

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

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

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

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