$x = \alpha(s), y = \beta(s)$
a)$\alpha'(s) = 3 \to \alpha = \int 3 ds = 3s + c_1$
$ \beta'(s)=-4 \to \alpha = (7-4)s + c_1 = (7 + \beta')s + c_1$
b)$\alpha'(s) = \alpha(s)^2 \to \alpha'(s)\beta(s)^2 = \alpha(s)^2\beta(s)^2$ (i)
$\beta'(s)=\beta(s)^2 \to \beta'(s)\alpha(s)^2 = \beta(s)^2\alpha(s)^2$ (ii)
(i)-(ii) = $\alpha'(s)\beta(s)^2 - \beta'(s)\alpha(s)^2 = 0$
I'm stuck. I need to find the equation of the line ax+by + c = 0. But I can't relate $x, y$ without the $s$.
EX from the book.
$2yu_x + u_y = (2y^2 + x)sin(2xy) \in \mathbb R$
$\alpha'(s) = 2\beta(s)$
$\beta'(s) = 1 \to \int \beta'(s) ds = \int 1 ds = \beta(s) = s + c_1$
$\int \alpha'(s) ds = \int 2(\beta) ds = \int 2s ds + \int 2c_1 ds$
$\alpha(s) = s^2 + 2sc_1 + c_2 = (s + c_1)^2 + c_2 - c_1^2$
$x = y^2 + c$
Thanks.
$$\boxed{3u_x-4u_y=x^2}$$ The Charpit_Lagrange characteristic ODEs are : $$\frac{dx}{3}=\frac{dy}{-4}=\frac{du}{x^2}$$ A first characteristic equation comes from solving $\frac{dx}{3}=\frac{dy}{-4}$ : $$4x+3y=c_1$$ A second characteristic equation comes from solving $\frac{dx}{3}=\frac{du}{x^2}$ : $$u-\frac19 x^3=c_2$$ The general solution of the PDE (without taking account of boundary condition) on form of implicite equation $c_2=F(c_1)$ is : $$u-\frac19 x^3=F(4x+3y)$$ $F$ is an arbitrary function (to be determined according to some boundary condition). $$\boxed{u(x,y)=\frac19 x^3+F(4x+3y)}$$ Since no boundary condition is specified in the wording of the question the function $F$ cannot be determined. This means that they are infinity many solutions because infinity many functions $F$ can be considerted.
Checking of the above result :
$u_x=\frac13 x^2+4\frac{dF}{dX}$ with $X=4x+3y$.
$u_y=3\frac{dF}{dX}$ with $X=4x+3y$.
$3u_x-4u_y=3\left(\frac13 x^2+4\frac{dF(X)}{dX}\right)-4\left(3\frac{dF(X)}{dX}\right)$
After simplification : $3u_x-4u_y=x^2$. Thus the PDE is satisfied any $F$.
SECOND PROBLEM :
$$\boxed{x^2u_x + y^2u_y = x^3}$$
The Charpit_Lagrange characteristic ODEs are : $$\frac{dx}{x^2}=\frac{dy}{y^2}=\frac{du}{x^3}$$ A first characteristic equation comes from solving $\frac{dx}{x^2}=\frac{dy}{y^2}$ : $$\frac{1}{y}-\frac{1}{x}=c_1$$ A second characteristic equation comes from solving $\frac{dx}{x^2}=\frac{du}{x^3}$ : $$u-\frac12 x^2=c_2$$ The general solution of the PDE (without taking account of boundary condition) on form of implicite equation $c_2=F(c_1)$ is : $$u-\frac12 x^2=F(\frac{1}{y}-\frac{1}{x})$$ $F$ is an arbitrary function (to be determined according to some boundary condition). $$\boxed{u(x,y)=\frac12 x^2+F(\frac{1}{y}-\frac{1}{x})}$$ Since no boundary condition is specified in the wording of the question the function $F$ cannot be determined. This means that they are infinity many solutions because infinity many functions $F$ can be considerted.
Checking of the above result :
$u_x=x+\frac{1}{x^2}\frac{dF}{dX}$ with $X=\frac{1}{y}-\frac{1}{x}$.
$u_y=-\frac{1}{y^2}\frac{dF}{dX}$ with $X=\frac{1}{y}-\frac{1}{x}$.
$x^2u_x + y^2u_y =x^2\left(x+\frac{1}{x^2}\frac{dF}{dX}\right)+y^2\left(-\frac{1}{y^2}\frac{dF}{dX}\right)$
After simplification : $x^2u_x + y^2u_y=x^3$. Thus the PDE is satisfied any $F$.