What Shape Do the Characteristic Curves of This PDE Form?

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SUMMARY

The characteristic curves of the partial differential equation (PDE) given by -4y∂u/∂x + ∂u/∂y = 0 are derived through the method of characteristics. The solution shows that y is a linear function of t, while x is a quadratic function of t, specifically x = -2t² - 4y₀t + x₀. Consequently, the relationship between x and y indicates that the characteristic curves form parabolas in the xy-plane.

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Homework Statement


suppose u(x,y) satisfies the partial differential equation:
-4y\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}=0
Find the characteristic curves for this equation and name the shape they form

The Attempt at a Solution


\frac{dy}{dt}=1 \Longrightarrow y=t+y_0
\frac{dx}{dt}=-4y=-4(t+y_0) \Longrightarrow x=-2t^2-4y_0t+x_0

What can I say about the shape these curves form?
 
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jimmycricket said:

Homework Statement


suppose u(x,y) satisfies the partial differential equation:
-4y\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}=0
Find the characteristic curves for this equation and name the shape they form



The Attempt at a Solution


\frac{dy}{dt}=1 \Longrightarrow y=t+y_0
\frac{dx}{dt}=-4y=-4(t+y_0) \Longrightarrow x=-2t^2-4y_0t+x_0

What can I say about the shape these curves form?

y is a first-order polynomial in t. x is a quadratic in t. Therefore x is a quadratic in y. What shape does the graph of a quadratic function have?
 

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