Answer to Question #215403 in Differential Equations for Asmita

Question

Obtain the characteristic curves for the partial differential equation 4uxx+5uxy+uyy+ux+uy=0.

Answer 1

Given Partial differential equation that is -

$=4u_{xx}+5u_{xy}+u_{yy}+u_{x}+u_{y}=0$

Now comparing this equation with the general form we get , which is -

$Au_{xx}+Bu_{xy}+Cu_{yy}+Du_{x}+Eu_{y}+Fu=G$

Now comparing both the equation we get ,

$A=4,B=5,C=1,D=1,E=1,F=0$

now taking out discriminant , which is equal to -

$B^{2}-4AC=25-4\times4\times1=9>0$

so the given characteristics curve will be hyperbolic .

Corresponding characteristics curve equation will be given as -

$\dfrac{dy}{dx}=\dfrac{-\xi_{x}}{\xi_{y}}$ $=-(\dfrac{-B+\sqrt{B^{2}-4AC}}{2A})=-\dfrac{(-5+9)}{8}=-\dfrac{1}{2}$

$\dfrac{dy}{dx}=\dfrac{-\eta_{x}}{\eta_{y}}$ $=-(\dfrac{-B-\sqrt{B^{2}-4AC}}{2A})=\dfrac{14}{8}=\dfrac{7}{4}$

To find ${\xi}$ and ${\eta}$ , we solve the above two equations for $y$ , we get -

$y=\dfrac{-1}{2}x+c_{1}$ and $y=\dfrac{7}{4}x+c_{2}$

Which give the value of two constants $c_{1}\ and \ c_{2}$ as given below-

$c_{1}=y+\dfrac{1x}{2}$ and $c_{2}=y-\dfrac{7x}{4}$

${\therefore} {\xi}(x,y)=y+\dfrac{x}{2}=c_{1}$ and $\eta({x,y})=y-\dfrac{7x}{4}=c_{2}$

These are equations of straight lines known as characteristics lines for given hyperbolic equation .