"How to find first integral in a given..." was answered by our community

Deven Livingston

Answered question

2022-04-23

How to find first integral in a given quasilinear partial differential equation?

y (x + y) \frac{\partial z}{\partial x} + x (x + y) \frac{\partial z}{\partial y} = 2 (x^{2} - y^{2}) + x z - y z .

Jocelynn Meyer

Beginner2022-04-24Added 9 answers

Step 1

y (x + y) \frac{\partial z}{\partial x} + x (x + y) \frac{\partial z}{\partial y} = 2 (x^{2} - y^{2}) + x z - y z .

The wrote:

\frac{d x}{y (x + y)} = \frac{d y}{x (x + y)} = \frac{d z}{2 (x^{2} - y^{2}) + x z - y z} .

The correctly found a first characteristic equation

x^{2} - y^{2} = C_{1}

For a second characteristic equation :

\frac{d x}{y (x + y)}

= \frac{d y}{x (x + y)}

= \frac{d x - d y}{y (x + y) - x (x + y)}

= \frac{d z}{2 (x^{2} - y^{2}) + x z - y z}

\frac{d x - d y}{- (x^{2} - y^{2})} = \frac{d z}{2 (x^{2} - y^{2}) + (x - y) z}

\frac{d x - d y}{- C_{1}} = \frac{d z}{2 C_{1} + (x - y) z}

With

t = x - y

- \frac{d t}{C_{1}} = \frac{d z}{2 C_{1} + t, z}

C_{1} \frac{d z}{d t} + t, z + 2 C_{1} = 0

This is a linear first order ODE which solution is :

e^{\frac{t^{2}}{2 C_{1}}} z + \sqrt{2 π C_{1}} erfi (\frac{t}{\pm \sqrt{2 C_{1}}}) = C_{2}

Fonction erfi :

e^{\frac{{(x - y)}^{2}}{2 (x^{2} - y^{2})}} z + \sqrt{2 π (x^{2} - y^{2})} erfi (\frac{x - y}{\pm \sqrt{2 (x^{2} - y^{2})}}) = C_{2}

e^{\frac{(x - y)}{2 (x + y)}} z \pm \sqrt{2 π (x^{2} - y^{2})} erfi (\sqrt{\frac{x - y}{2 (x + y)}}) = C_{2}

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