Answer to Question #200112 in Differential Equations for Kevin

Question

3) find the integral surface of the equation (y-z){2xyp+(x^2-y^2)q}+z(x^2-y^2)=0 through the curve x=t^2, y=0 , z=t^3

Answer 1

$Question:\\ find \space the\space integral\space surface \space of \space the\space equation \\(y-z){2xyp+(x^2-y^2)q}+z(x^2-y^2)=0\\ through\space the\space curve\\ x=t^2, y=0 , z=t^3\\ solution: \\ use\space lagrange's \space method \\ \frac{dx}{(y-z)2xy}=\frac{dy}{(y-z)(x^2-y^2)}=\frac{dz}{y^2-x^2}\\ Now \space I=II\\ \frac{dx}{(y-z)2xy}=\frac{dy}{(y-z)(x^2-y^2)}\\ {(x^2-y^2)}{dx}-{2xy}{dy}=0\\ this \space is\space exact\space equation\space\\ (since\space if\space Mdx+Ndy=0,\space and\space \frac{∂M}{∂y}=\frac{∂N}{∂x}\space then\space it \space is \space called\space exact \space equation. Now\space we \space know\space that\\ for\space exact \space equations\space ,solution \space is\space\\ \int Mdx+\int [N-\frac{∂\int Mdx}{∂y}] dy=c_1\\ hence\\ \frac{x^3}{3}-y^2x=c_1.................................(1)\\ Now \space II=III\\ dy=(z-y)dz\\ \frac{dy}{dz}+y=z\\ it \space is \space linear \space diff.\space eq.// for \space it's \space solution\space \\ I.F.=e^{\int P dz}, \space where \space p=1\space multiple\space of \space y\\ I.F.=e^{\int 1 dz}={e}^{z}\\ and \\ solution\space as\\ {e}^{z}.y=\int {e}^{z}.z \space dz {e}^{z} .y= {e}^{z}(z-1)+c_2..............................(2)\\ now\space we \space have\space 2\space equations\\ \frac{x^3}{3}-y^2x=c_1.................................(1)\\ {e}^{z} .y= {e}^{z}(z-1)+c_2..............................(2)\\ given \space curve x=t^2, y=0 , z=t^3; put \space this\space values\space in\space equations; \frac{(t^2)^3}{3}-(0)^2(t^2)=c_1\\ c_1=\frac{(t^2)^3}{3}................................(4)\\ and\\ {e}^{z} .y= {e}^{z}(z-1)+c_2\\ {e}^{t^3} .(0)= {e}^{t^3}(t^3-1)+c_2\\ c_2={e}^{t^3}(1-t^3)...........................(5)\\ now\space put\space value\space of \space t^3=\sqrt{3c_1}\space in\space (5)\\ c_2={e}^{\sqrt{3c_1}}(1-\sqrt{3c_1})\\ now\space put\space value\space of \space c_1\space and\space c_2\\ {e}^{z} .y- {e}^{z}(z-1)={e}^{\sqrt{3(\frac{x^3}{3}-y^2x)}}(1-\sqrt{3(\frac{x^3}{3}-y^2x)})\\$